You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Physics

Work Energy Theorem Application

The work done...

Question

The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is

4.36 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.17 kJ

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8.91 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9.82 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
5.17 kJ

$W = m g s i n θ \times s$

$= 2 \times 10^{3} \times s i n 15^{\circ} \times 10$

$= 5.17 k J$

Suggest Corrections

Similar questions

Q. The work done in slowly pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of

15^{\circ}

with the horizontal by a force parallel to the incline is:

Q. The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of

15^{\circ}

with the horizontal is

The work done in pulling up a block of wood weighing 2kN for a length of 10 m on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is $(s i n 15^{\circ} = 0.259)$

Q. How much is the work done in pulling up a block of wood weighing

2 k N

for a length of

10 m

on a smooth plane inclined at an angle of

30^{o}

with the horizontal?

Q. The work done in pulling up a block of wood weighing

2 k N

through a distance length of

10 m

on a smooth plane inclined at an angle of

15^{o}

with the horizontal is

(

sin 15^{o} = 0.2588

)

Join BYJU'S Learning Program

The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of 15∘ with the horizontal is

The correct option is B 5.17 kJ W=mg sinθ×s =2×103×sin15∘×10 =5.17kJ

The work done in pulling up a block of wood weighing 2 kN for a length of 10m on a smooth plane inclined at an angle of $15^{\circ}$ with the horizontal is

The correct option is B
5.17 kJ

$W = m g s i n θ \times s$

$= 2 \times 10^{3} \times s i n 15^{\circ} \times 10$

$= 5.17 k J$