## Newtonâ€™s Second Law Practice Problems

Here are Newtonâ€™s second law practice problems to help you master the calculations involving the second law of motion.

### Problem 1

An unbalanced force of 20 Newtons acts on a 4.0-kilogram mass. What acceleration does it give it?

**Data**

Force acting on the body, f = 20 N

Mass of the body, m = 4 kg

**Unknown **

Acceleration of the body, a = ?

**Formula**

The formula for Newtonâ€™s second law of motion is F = ma

and to find the acceleration, we need to make a subject of the formula

a = f/m

We will apply the above formula to solve the problem

**Solution**

Substitute your data into the formula

a = 20/4 = 5 ms^{-2} [ Because the unit of acceleration is ms^{-2} ]

**Therefore, the acceleration of the body is 5 meters per second square.**

### Problem 2

A 900 grams stone is pushed along a tarmac by a horizontal force of 20 Newton. A frictional force of 8 newtons opposes the motion. What is the acceleration given to the stone?

**Data **

Mass of the stone, m = 900 g = (900/1000) kg = 0.9 kg

Horizontal force, f_{h} = 20 N

Frictional force, f_{r } = 8 N

**Unknown**

Unbalanced force, F = f_{h} â€“ f_{r} = 20 â€“ 8 = 12 N

The acceleration, a = ?

**Formula **

Force = mass x acceleration

Which can be written as

F = m x a

**Solution**

Now we can insert our data into the formula

The acceleration, a = F/m = 12/0.9 = 13.3 ms^{-2}

**Therefore, the acceleration of the stone is 13 ms ^{-2}**

### Problem 3

A ball of mass 0.3 kilograms moving at a velocity of 20 meters per second is suddenly hit by a force of 5 newtons for a time of 0.03 seconds. Find its new velocity of motion.

**Data**

Mass of the ball, m = 0.3 kg

Initial velocity, u = 20 m/s

Force, F = 5 N

Time, t = 0.03 s

**Unknown**

Final velocity, v = ?

**Formula**

Since F = ma

and a = (v-u)/t

We can see that

F = m(v-u)/t = (mv â€“ mu)/t

We can make v subject of the formula from F = m(v-u)/t

F/m = (v-u)/t

and now

Ft/m = v â€“ u

Therefore, the formula we need to apply is

v = (Ft/m) + u

**Solution**

It is now time for us to put our data into the formula

v = (5 x 0.03/0.3) + 20

Now, our final velocity, v = (0.15/0.3) + 20 = 0.5 + 20 = 20.5 m/s

**Therefore, our final velocity is 20.5 meters per second**

### Problem 4

A body of mass 0.1 kilograms dropped from a height of 8 meters onto a hard floor and bounces back to a height of 2 meters. Calculate the change of momentum. If the body is in contact with the floor for 0.1 seconds, what is the force exerted on the body? ( Take gravitational acceleration, g = 10 ms^{-2}).

**Data **

Mass of the body, m = 0.1 kg

Before hitting the floor, the height h_{1} = 8 m

After hitting the floor, the height of bouncing h_{2} = 2 m

Gravitational acceleration, g = 10 ms^{-2}

Time, t = 0.1 s

Before hitting the floor, the initial velocity, u_{1} = 0

After hitting the floor, the final velocity after bouncing, v_{2} = 0

**Unknown**

Before hitting the floor, the final velocity, v_{1} = ?

After hitting the floor, the initial velocity after bouncing, u_{2} = ?

**Formula**

The first formula

v_{1}^{2} = u_{1}^{2} + 2gh_{1}

The second formula

v_{2}^{2} = u_{2}^{2} â€“ 2gh_{2}

The final formula to apply is

F = (mv_{1} â€“ (-mu_{2}))/t = (mv_{1} + mu_{2})/t [For the body bouncing and moving away from the floor, the momentum is negative]

**Solution**

We can apply our data to the first formula

v_{1}^{2} = u_{1}^{2} +2gh_{1} = u_{1}^{2} +2gh_{1} = 0 + 2 x 10 x 8 = 160

Therefore

v_{1}^{2} = 160

and v_{1} = âˆš160 = 12.7 m/s = 13 m/s

Applying our data to the second formula, we will have

v_{2}^{2} = u_{2}^{2} â€“ 2gh_{2}

0 = u_{2}^{2} â€“ 2 x 10 x 2

u_{2}^{2} â€“ 40 = 0

We can now collect like terms to obtain

u_{2}^{2} = 40

u_{2} = âˆš40 = 6.325 m/s = 6 m/s

Applying the final formula, we can now find the force exerted on the body

F = (mv_{1} + mu_{2})/t

substituting our data and the rest of the values, we will get

Force, F = (mv_{1} + mu_{2})/t = m(v_{1} + u_{2})/t = 0.1(13 + 6)/0.1 = 19 N

**Therefore, the force exerted on the body is 19 Newton**

### Problem 5: Newtonâ€™s Second Law Practice Problems

A body of mass 2 kilograms falls freely from rest, through a height of 50 meters, and comes to rest having penetrated 5.0 centimeters of sand.

a. Calculate the velocity with which the body hits the sand

b. The time taken in falling

c. The average force exerted by the sand in bringing the body to rest.

**Data**

Mass of the body, m = 2 kg

Initial height, h_{1} = 50 m

Final height, h_{2} = 5 cm = 5/100 m = 0.05 m

Gravitational acceleration, g = 10 ms^{-2}

Initial velocity, u = 0

**Unknown**

v = ?

t = ?

retardation, a = ?

F = ?

**Formula**

The first formula is

v^{2} = u^{2} + 2gs

The second formula to apply is

v = u + gt

We will also apply v^{2} = u^{2} + 2as to find retardation

And then finally apply the formula for Newtonâ€™s second law of motion to solve the force exerted by the sand on the body

F = ma

**Solutions**

a. We can add our data into the equation v^{2} = u^{2} + 2gs

Hence,

v^{2} = 0 + 2 x 10 x 50 = 1000

v = âˆš1000 = 31.623 m/s = 31 m/s

b. Substitute your data into the second formula

v = u + gt

we make t subject of the formula

v = gt [since u = 0]

Thus,

t = v/g = 31/10 = 3.1 s

c. We need to find the retardation by applying the formula v^{2} = u^{2} + 2as before we can go ahead to find the force

Insert our data into v^{2} = u^{2} + 2as

**Note: After hitting the sand, our initial velocity = 31 m/s and the final velocity = 0 **

Hence,

0 = 31^{2} + 2 x a x 0.05

We make a subject of the formula

a = -(961/0.1) = -9610 m/s^{2} [The negative sign indicates negative acceleration or retardation]

Therefore, a = 9610 m/s^{2}

We can now comfortably apply our final formula that says Force (F) = Mass(m) x Acceleration (a)

Thus,

F = ma = 2 x 9610 = 19,220 N

**Therefore, the force exerted by the sand in bringing the body to rest is 19,220 Newton**

### Problem 6: Newtonâ€™s Second Law Practice Problems

A towing van tows a car of mass 1,500 kilograms along a level road and accelerates at 0.8 meters per second square. What is the tension in the towing string?

If the towing string suddenly breaks when the car reaches a speed of 20 meters per second, calculate the distance covered by the car before it comes to rest on the application of a braking force of 2000 newtons.

**Data **

Mass of the car, m = 1,500 kg

Acceleration, a = 0.8 ms^{-2}

Initial speed, u = 20 m/s

Final speed, v = 0 [Because the car finally came to rest, then its final velocity is zero]

Braking force = 2000 N

**Unknown**

The tension of the string, F = ?

Change in momentum = ?

Time taken, t = ?

Deceleration, a_{D} = ?

Distance covered, s = ?

**Formula**

To find the tension of the string, we will apply

F = ma

For change in momentum, the formula will be

Change in momentum = mu â€“ mv

The to find time taken is

t = change in momentum / tension on the string (F)

Finding deceleration, we will apply

a = (v â€“ u) / t

And finally, the last formula to find the distance covered, we use

s = (v^{2} â€“ u^{2}) / 2a

**Solution**

Let us find the tension on the string by substituting our data into F = ma

F = 1,500 x 0.8 = 1,200 N

The change in momentum = mv â€“ mu = m(v â€“ u) = 1,500 (20 â€“ 0) = 1,500 x 20 = 30,000 kgms^{-1}

To calculate the time taken, we will insert our data into the formula

Time taken, t = change in momentum / tension on the string = 30,000 / 1,200 = 25 seconds

The deceleration will be

a_{D} = (v â€“ u) / t = (0 â€“ 20) / 25 = â€“ 0.8 ms^{-2}

Now, we can go ahead to calculate the distance covered

s = (v^{2} â€“ u^{2}) / 2a_{D} = (0^{2} â€“ 20^{2}) / 2 x (-0.8) = -400/-1.6 = 250 m

**Therefore, the distance covered is 250 meters**

Therefore, I hope these Newtonâ€™s second law practice problems have helped you understand how to solve problems related to the second law of motion.

You can share your biggest challenge in Newtonâ€™s second law practice problems so that i can help you solve them.

*You may also like to read:*

Newtonâ€™s Second Law of Motion Examples

15 Newtonâ€™s Laws of Motion Project Ideas

*Sources:*