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Force Numericals

Last updated on July 5th, 2023 at 03:21 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.

Formulas Used to Solve Force Numericals

F = ma ………………………… [1]
F = m [v-u]/t………………. [2]
F = [mv-mu]/t…………….. [3]
Ft = [mv-mu]
Impulse = change in momentum ……………[4]
F = Impulse/t ……………………[5]

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

Force Numericals – with solution

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

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 Ns

Force 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.

See also  Numerical problem on Terminal Velocity with solution

Solution:

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

v=0

s= 100 m

Using the equation of motion: v2 = u2 – 2as
=> 0 = 100 – 2.a.100
=> a = 100/[200] = 0.5 m/s2

acceleration a = 0.5 m/s2

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.

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