WORK ENERGY AND POWER

1 )

How much is the work done in pulling up a block of wood weighing 2KN for a length of 10m on a smooth plane inclined at an angle of 30° with the horizontal ?

1.732 KJ

17.32 KJ

10 KJ

100 KJ

View Solution

Answer :

Workdone = F × d × cos θ

θ = 30°

d = 10 m

F = 2 KN

W = 2 × 10³ × 10 × cos 30

= 17.32 × 10³ J

= 17.32 KJ

2 )

A force of 7N, making an angle θ with the horizontal, acting on an object displaces it by 0.5m along the horizontal direction. If the object gains K.E. of 2J, what is the horizontal component of the force?

14N

View Solution

Answer :

Kinetic energy = Fd cosθ

2 = 7 × 0.5 × cosθ

cosθ = 0.571

Horizontal component = F cosθ = 7 × 0.571 = 3.99 N

3 )

A 60 kg JATAN with 10 kg load on his head climbs 25 steps of 0.20m height each. What is the work done in climbing? (g = 10 m/s²)

3500 J

350 J

35 J

4000 J

View Solution

Answer :

W = F × d

= M × g × d

here M = 60 + 10 = 70 kg

W = 70 × 10 × d

W = 700d

d = no of steps × height of each step

= 25 × 0.2 = 5 m

W = 700 × 5

= 3500 J

4 )

A ball of mass 5 kg is striding on a plane with initial velocity of 10 m/s. If co-efficient of friction between surface and ball is (1/2), then before stopping it will describe ……. (g = 10 m/s²)

15 m

17 m

7 m

10 m

View Solution

Answer :

m = 5 kg

V_o = 10 m/s

μ = (1/2)

V = 0 as it stops

V² = V_o² + 2 as

as surface is frictional hence V² = V_o² + 2 μgs

0 = 10² + {2 × (1/2) × 10 × s}

s = – 10 m hence displacement = 10 m

5 )

The relationship between force and position is shown in the figure given (in one dimensional case) calculate the work done by the force in displacing a body from x = 0 cm to x = 5 cm

Description: https://stemez.com/subjects/joint_entrance_examination/1rjeephysics/1rjeephysics/1RJeePhysics/1R04-0423_files/image002.jpg

30 ergs

70 ergs

20 ergs

60 ergs

View Solution

Answer :

W = area under curve of F – x graph till x = 5 cm

= [2 × 10] + [1 × 20] + [1 × 10] – [1 × 20]

= 20 + 20 + 10 – 20

= 30 ergs

Topic Categories