Question

Write whether the work done in the following cases are positive, negative or zero. (a) Work done by frictional force. (b) Work done by centripetal force on a body moving in a circular path. (c) Work done by gravitational force in a freely falling body. (d) Work done by a person carrying a load on his head and walks along a horizontal level road​

Answer 1

Answer:

a) Work done by frictional force

Ans : positive

The work done by frictional force will be positive when the displacement of the body is in the direction of the force. ... Since, the direction of motion of the block and the direction of frictional force is opposite to each other, the work done by friction is negative. The work done by friction can also be zero.

b) Work done by centripetal force on a body moving in a circular path

Ans : also zero

The work done on a body moving in a circular path is also zero. This is because, when a body moves in a circular path, then the centripetal force acts along the radius of the circle, and it is at right angles to the motion of the body. Thus, the work done in the case of moon moving round the earth is also zero.

c) Work done by gravitational force in a freely falling body

Ans : W=mgh

The work done by gravity is W=mgh.

d) Work done by a person carrying a load on his head and walks along a horizontal level road

Ans : Work done against gravity is zero as there is no displacement along or opposite to the direction of force of gravity.

Explanation:

Hope it's helpful to you

Answer 2

Work done by frictional force, centripetal force and Work done by a person carrying a load on his head and walking along a horizontal level roadis negative, zero, and zero respectively.

• Work done is the amount of energy required to move an object or lift a load to some distance or some angle.
• The formula of work done is:
• $fscosθ$
• Here, f is the force s is the distance traveled and θ is the angle between the force and the distance.
• The frictional force is in the opposite direction of the displacement so, the angle θ is 180°. Work done =
• $fscos(180) = - fs$
• The centrifugal force is perpendicular to the direction of the displacement so, the angle θ is 90°. Work done =
• $fs(cos90) = 0$
• Now, The force is in the perpendicular direction concerning the displacement when a person carrying a load on his head walks along a horizontal level road so, the angle θ is 90° because the force of gravity is in the vertical direction while displacement is horizontal. Work done =
• $fs(cos90) = 0$

#SPJ3