Question 1.

The engine of a vehicle moves it forward with a force of 9600N against a resistive force of 2200N. If the mass of the vehicle is 3400kg, calculate the acceleration produced.

Solution:

Note: the net force with which the vehicle accelerated is the difference between the two forces, i.e.  

9600N - 2200N = 7400 Newtons

Mass = 3400kg, acceleration = ?

Therefore, using the formula for force,

F = ma

7400 = 3400 x a

a = ⁷⁴⁰⁰/₃₄₀₀ = approx. 2.18m/s²

Question 2.

A net force of 15N acts upon a body of mass 3kg for 5s, calculate the change in the speed of the body.

Solution:

Given: Force = 15N, mass = 3kg, time = 5s, ∆v = ?

 Using the force formula, F= m x ^∆V/t      

Restating the formula, we have ∆v = ^Ft/_m

∆v = ^{15 x 5}/₃ = 25m/s

Question 3.

A bob of a simple pendulum has a mass of 0.02kg. Determine the weight of the bob. [g = 10ms^-2]

Solution:

Given: mass = 0.02kg, acceleration due to the force of gravity = 10ms^-2

Using the formula for force, F = mg, where g (acceleration due to the force of gravity replaces a (normal acceleration),

Weight, W = mg

Therefore, W = 0.02 x 10 = 0.2 Newtons

Question 4.

What change in velocity would be produced on a body of mass 4 kg if a constant force of 16 N acts on it for 2 s?

Solution:

Given: Force = 16 Newtons, mass = 4 kg, time = 2 s, change in velocity (∆v) = ?

Using the formula for force, F = m x ^∆V/t

16 = 4 x ^∆V/2

Restating the equation, ^{16 x 2}/₄ =∆v

∆v = 8 m/s

Question 5.

A force of 16 N applied to a 4.0kg block that is at rest on a smooth, horizontal surface. What is the velocity of the block at t = 5 seconds?

Solution:

Using the formula for force,

F = m x ^∆V/t  or F = m x  ^{(v - u)}/_t    

Given: m = 4.0 kg, final velocity v = ?, initial velocity u = 0, time = 5 seconds.

Therefore,

16 = 4.0 x ^{(v - 0)}/₅

16 = 4.0v/5

4.0v = 80

v = 80/4 = 20

Velocity of the block is 20 m/s

See more calculations of force based on Newton's second law of motion.