Last updated on May 3rd, 2023 at 04:33 pm

Newton’s second law of motion provides us with the formula of force. We will use different forms of this formula in this post to solve a set of **Numerical Problems on Force**.

These **Force Numericals** in this post are especially based only on the 2nd law of motion.

We have another post that includes Numericals on laws of motion and force for class 9.

## Force Numericals – with solution

**Formulas Used** to Solve Force Numericals

F = ma ………………………… [1]

F = force, m =mass, a = acceleration, v = final velocity, u = initial velocity, t = time duration, Ft = impulse, mv-mu = change in momentum

F = m [v-u]/t………………. [2]

F = [mv-mu]/t…………….. [3]

Ft = [mv-mu]

Impulse = change in momentum ……………[4]

F = Impulse/t ……………………[5]

Here is a set of numerical problems (with solution) based on the force formula derived using the Second Law of motion.

**Numerical Problems on Force** [with solution]

**1] A constant force acts on a body of mass 100 kg and produces in it an acceleration of 0.2 m s ^{−2}.Calculate the magnitude of the force acting on the body.**

**Solution:**

Given

Mass = 100 kg

a = 0.2 m s^{−2}F = ma

∴ Force =mass × acceleration

∴ F = (100 kg) × (0.2 m s^{−2}) = 20 N

Thus, the magnitude of the force acting on the body is 20 N.

**2] A cricket ball of mass 100 g is moving with a velocity of 10 m s ^{−1 }and is hit by a bat so that it turns back and moves with a velocity of 20 m s^{−1}. Find the impulse and the force if the force acts for 0.01 s.**

**Solution:**

Mass of the ball = 100 g = 0.1 kg

Initial velocity u and final velocity v are in opposite directions. Taking u as -ive and v as +ive we can write:

u =–10 m/s

v = 20 m/s

v-u = 20 – (-10) = 30 m/s ………………… [1]Impulse = Ft = force x time …………….[2]

We know, Ft= [mv-mu]

=>Impulse = m[v-u]

=>Impulse = 0.1 [30] = 3 NsForce F = Impulse/t

F = 3/0.01 N = 300 N

**3] A car moving at a speed of 36 km h−1 is brought to rest while covering a distance of 100 m. If the mass of the car is 400 kg, find the retarding force on the car and the time taken by the car to stop.**

Solution:

u = 36 km/h = 36x 5/18 m/s = 10 m/s

v=0

s= 100 m

Using the equation of motion: v

^{2}= u^{2}– 2as

=> 0 = 100 – 2.a.100

=> a = 100/[200] = 0.5 m/s^{2}acceleration a = 0.5 m/s

^{2}mass m = 400 kg

Force = mass x acceleration = 400 x 0.5 N = 200 N

We have another post that includes solved numericals based on all 3 laws of motion and force for class 9.