DC Pandey Solutions: Work, Energy & Power - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Introductory Exercise 6.1

Q.1. A block is pulled a distance x along a rough horizontal table by a horizontal string. If the tension in the string is T, the weight of the block is W, the normal reaction is N and frictional force is F. Write down expressions for the work done by each of these forces.
Ans. 7x, O, O, - Fx

Work done by



Work done by

Work done by

Work done by

Q.2. A particle is pulled a distance l up a rough plane inclined at an angle α to the horizontal by a string inclined at an angle β to the plane (α + β < 90°). If the tension in the string is T, the normal reaction between the particle and the plane is N, the frictional force is F and the weight of the particle is W. Write down expressions for the work done by each of these forces.
Ans. 

Work done by


Work done by

Work done by

= - Wl sin α
Work done by

Q.3. A bucket tied to a string is lowered at a constant acceleration of g/4. If the mass of the bucket is m and is lowered by a distance l then find the work done by the string on the bucket.
Ans: 

∴
 
Work done by string

Q.4. A 1.8 kg block is moved at constant speed over a surface for which coefficient of friction  It is pulled by a force F acting at 45° with horizontal as shown in figure. The block is displaced by 2 m. Find the work done on the block by (a) the force F (b) friction (c) gravity.

Ans. (a) 7.2 j  (b) - 7.2 J (c) zero

μN = F cos 45°              …(i)


Substituting value of N from Eq. (ii) in Eq. (i).

Work done by force

Work done by friction

Work done by gravity

Q.5. A small block of mass 1 kg is kept on a rough inclined wedge of inclination 45° fixed in an elevator. The elevator goes up with a uniform velocity v = 2 m/s and the block does not slide on the wedge. Find the work done by the force of friction on the block in 1 s. (g = 10 m/s²)

Ans 10 J

F = mg sin 45°


Displacement of lift in 1s = 2 m
Work done  by force of friction

Q.6. Two equal masses are attached to the two ends of a spring of force constant k. The masses are pulled out symmetrically to stretch the spring by a length 2x₀ over its natural length. Find the work done by the spring on each mass.
Ans. 

Total work-done by spring on both masses

= PE of the spring when stretched by 2x₀

∴  Work done by spring on each mass

Q.7. Force acting on a particle varies with displacement as shown in figure. Find the work done by this force on the particle from x = - 4 m t o x = + 4 m .

Ans: 30J

Work done = Area under the curve

= A₁ + A₂ + A₃ + A₄
 
= 30 Nm