Q.

A box of mass 50kg at rest is pulled up on an inclined plane 12m long and 2m high by a constant force of 100N applied parallel to the inclined plane. When it reaches the top of the inclined plane if its velocity is 2ms–1, the work done against friction in Joules is (g = 10ms–2)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

50

b

100

c

150

d

200

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

WF = Work done by the applied force = 10 x 12 =1200JΔV = Change in P.E = mgh = 50 x 10 x 2 = 1000JΔK = Change in K.E. = 1/2 x 50 x 22 = 100JNow, WF = ΔK + ΔV + lost energy Lost energy = (1200-100-1000)J = 100J.
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon
A box of mass 50kg at rest is pulled up on an inclined plane 12m long and 2m high by a constant force of 100N applied parallel to the inclined plane. When it reaches the top of the inclined plane if its velocity is 2ms–1, the work done against friction in Joules is (g = 10ms–2)