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Calculating Force Based On Newton's Second Law Of Motion (3)

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Question 1.

A block of mass 2.0 kg resting on a smooth horizontal plane is acted upon simultaneously by two forces, 10 N due North and 10 N due East. The magnitude of the acceleration produced by the forces on the block is?

Solution:

Using the formula for force, F = ma

Given, force = ?, mass = 2.0 kg, acceleration = ?

First, find the net force of the two acting on the mass. Using pythagora's theorem, the resultant force, R is given as

R2 = 102 + 102

R  = (102 + 102)    =  200  = 14.4 N

Applying the net force to the force formula,

F = ma

14.14 = 2.0 x a

a = 14.14/2 = 7.07 m/s2

Question 2.

A 0.05kg bullet travelling at 500 ms-1 horizontally strikes a thick vertical wall. It stops after penetrating through the wall a horizontal distance of 0.25 m. What is the magnitude of the average force the wall exerts on the bullet?

Solution:

The force the wall exerted on the bullet in bringing it to rest or stop equals the force the bullet was travelling with.

Using the formula for force, F = ma

Given: mass = 0.05 kg, initial velocity u = 500 ms-1, final velocity v = 0, distance travelled = 0.25 m, acceleration = ?

To find the acceleration of the bullet, use the equation of motion,

v2 = u2 + 2as, where v = final velocity, u = initial velocity,     a = acceleration,    s = distance moved.

Therefore, from the equation of motion,

a = (v2 - u2)/2s

substituting (v2 - u2/2s ) for a in the force equation, we have

F = m (v2 - u2/2s)

F = 0.05 ((0 - 5002)/2 x 0.25)

= 25 000 N

The force the wall exerted on the bullet is 25 000 Newtons.

Question 3.

When taking a penalty kick, a footballer applies a force of 30.0 N for a period of 0.05s. If the mass of the ball is 0.075 kg, calculate the speed with which the ball moves off.

Solution:

Using the formula for force,

F = m (v - u/t)

Given: Force = 30 N, mass = 0.075 kg, initial velocity,   u = 0, final velocity, v = ?, time t = 0.05 s

Therefore, with the above force formula,

30 = 0.075 (v - 0/0.05)

0.075 v = 30 x 0.05

v = 30 x 0.05/0.075 = 20

The speed (or velocity with which the ball moves off is  20 m/s

Question 4.

A body of mass 2 kg moving vertically upwards has its velocity increased uniformly from 10 ms-1 to 40 ms-1 in 4s. Neglecting air resistance, calculate the upward vertical force acting on the body, taking acceleration due to gravity as 10 ms-2.

Solution:

Neglecting air resistance, two forces act on the body as it moves upwards: (1) the upward vertical force, F, and (2) the downward force due to gravity, mg.

The net force is the difference between the two forces.

i.e.

F - mg = ma

F = ma + mg

F = m (a + g)

Notice that  a = Dv/t

= v - u/t

= 40 - 10/ 4

= 30/4 = 7.5 ms-2

Therefore, from

F = m (a + g)

F = 2 (7.5 + 10)

F = 2 x 17.5

F = 35 N

The upward vertical force acting on the body is 35 Newtons

Question 5.

A force on a body causes a change in the momentum of the body from 12 kgms-1 to 16 kgms-1 in 0.2s. Calculate the magnitude of the impulse.

Solution:

From Newton's second law of motion

F = (mv - mu)/t

Ft = mv - mu

Ft = Impulse (I)

Therefore, I = mv - mu

Where mv = final momentum, mu = initial momentum.

I = 16 - 12 = 4 Ns

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