Question

1.a) If you lift a backpack up, does the force exerted by you on the backpack do positive, negative, or zero work? Explain your reasoning.

b) If you lower lower a backpack, does the force exerted by you on the backpack do positive, negative, or zero work? Explain your reasoning.

c) If you move a backpack horizontally with constant velocity, does the force exerted by you on the backpack do positive, negative, or zero work? Explain your reasoning.

2.

A parachutist jumps from an airplane that is at 1 km height above the ground. What is the magnitude of the work done by the air drag force if she lands with a terminal velocity of 5 m/s? The mass of the parachutist with the parachute is 95 kg.

The energy lost, Q =

Calculate the percentage from the initial potential energy that went to the final kinetic energy.

KEf/PEi⋅100%=

3.

A 2‑kg cannonball is fired straight up with an initial velocity of 49 m/s. Taking the initial position of the ball as the reference level (with zero height and zero potential energy), find the potential, kinetic and mechanical energies of the ball at different heights. Neglect the air resistance.

Height, in m	PE, in J	KE, in J	E, in J
h1 = 0
h2 = 10
h3 = 20
h4 = 30

How would you describe the energy transformation while the cannon ball is moving up?

Answer 1

1.

(a). work done, W = force x displacement

The work done while lifting a backpack up is negative. A force equal to the weight need to be applied to lift an object up and the motion is against gravity. The work done is negative as its motion is against gravity.

(b) The work done while lowering a backpack up is positive. The   force of gravity does the work and hence it is positive.

( c) the net force acting on the pack moving horizontally with uniform speed is zero. The gravity does no work as the motion of the pack is perpendicular to the force of gravity. Hence, the work done is zero .

2.

Potential energy of the parachutist at the height of 1 km is U = mgh = 95 x 9.8 x1000 = 931000 J

Kinetic energy of the parachutist when she lands, K = ½ m V² = ½ x 95 x 5 x 5 = 1187.5 J

Work done by the air drag, W = U – K = 931000 – 1187.5 = 929812.5 J

Work done by the air drag, W = 929812.5 J

Percentage of energy lost = ( U- K) x 100/U = 99.8 %