Force And Laws Of Motion Short Answer Question And Answers

Force And Laws Of Motion Short Answer Type Question And Answers

Question 1. Aforce of 8000 N acts on a truck, initially at rest for 10 s. Find

  1. velocity of the truck
  2. distance travelled by the truck
  3. Mass of the truck = 2000 kg

Answer.

∵ F = ma

8000 = 2000a

∴ a = \(\frac{8000}{2000}=4 \mathrm{~ms}^{-2}\)

(1) ∵ v = u + at

∴ v = 0 + 4 × 10 = 40 ms-1

(2) ∵ s = \(u t+\frac{1}{2} a t^2\)

= \(0+\frac{1}{2} \times 4 \times(10)^2\)

∴ S = 200 m.

Question 2. A ball weighing 500 g is lying on a floor. A player exerts a force of 250 N on the ball for a small time interval of 0.02 s. Find the velocity acquired by the ball.
Answer.

Given u = 0

m = 500 g = 0.5 kg

F = 250 N

t = 0.02 s

v = ?

We know that

F = ma = \(m\left(\frac{v-u}{t}\right)\)

⇒ 250 = \(0.5\left(\frac{v-0}{0.02}\right)\)

∴ v = \(\frac{250 \times 0.02}{0.5}=\frac{250 \times 2}{50}=10 \mathrm{~m} / \mathrm{s}\)

Question 3. A book weighing 800 g is kept on a horizontal table. How much force is exerted by the table on the book? (Take g = 10 ms−2)
Answer.

NEET Foundation Physics Force And Laws Of Motion Force Equal To Weight

The book exerts a force equal to its weight on the table.

This force is called ‘action’.

∴ Action = W = mg

= 0.8 × 10

= 8 N

According to Newton’s III law of motion, the table exerts an equal and opposite force on the book. It is called reaction.

∴ R = Action

= 8 N

It acts vertically upwards.

Question 4. Figure below show a trolley of mass 300 g lying on a smooth horizontal surface. A boy pulls it with a force 1.2 N. If a constant frictional force of 0.3 N acts on it, how much distance it will travel in 0.8 s?

NEET Foundation Physics Force And Laws Of Motion Frictional Force 1

Answer.

Since the two forces are acting in opposite direction, net force acting on the trolley

= F1 − F2

= 1.2 N − 0.3 N

= 0.9 N

We know that Newton’s II law of motion gives us

F = ma

0.9 N = 0.3 kg × a

∴ a = \(\frac{0.9 \mathrm{~N}}{0.3 \mathrm{~kg}}\)

= 3 ms-2

∵ S = \(u t+\frac{1}{2} a t^2\)

= \(0+\frac{1}{2} \times 3 \times(0.8)^2\)

= \(\frac{1}{2} \times 3 \times 0.8 \times 0.8\)

∴ S = 0.96 m

Question 5. In a cathode ray tube shown in figure below, an electron starts from cathode c. What will be its speed on reaching the anode, if an electric force of N acts on it?

  1. (Mass of electron = 9 × 10-31 kg)
  2. (Distance between cathode C and anode A is cm)

 

Answer.

By Newton’s II law of motion,

F = ma

1.8 × 10 = (9 × 10-31 kg) a

⇒ 9 × 10-31 a = 1.8 × 10-15

∴ a = \(\frac{1.8 \times 10^{-15}}{\left(9 \times 10^{-31}\right)}\)

∴ a = 2 × 1015 ms-2

We know that,

v2 = u + 2as

0 = 2 × 2 × 1015 × 0.4

⇒ v2 = 1.6 × 1015

⇒ v2 = 16 × 1014

∴ v = 4 × 107 ms-1

Leave a Comment