When you jump, you start by crouching down a bit, which lowers your center of ma

Question

When you jump, you start by crouching down a bit, which lowers your center of mass to a distance d 40 cm below where it normally is. You then exert a downward contact force on the ground equal to about 2.5 times your weight (mg); we may assume that this force is constant while it acts. After your center of mass rises past its usual location, you leave the ground with an initial velocity v, and you reach a maximum height h. Ignore air resistance for this problem.

a) Draw a free-body diagram for your body during the interval while you are pushing against the ground with a force of 2.5mg. What is the net force on your body during that time?

b) What is the work done on the Earth by the force of your feet pushing on the Earth during that time?

Explanation / Answer

a) Here weight acts downward and norma force acts upward.

Fnet = N - weight

= 2.5*m*g - m*g

= 1.5*m*g

Net workdone = gain inkinetic energy

Fnet*h = 0.5*m*v^2

1.5*m*g*0.4 = 0.5*m*v^2

==> 0.6*g = 0.5*v^2

v = sqrt(1.2*g)

= sqrt(1.2*9.8)

= 3.43 m/s

b) the work done on the Earth by the force of your feet pushing on the Earth during that time is Zero

Beacuse, Workdone = F*d (here F is force and d is displacement)

here displacement of earth is zero.

so, Workdone = 0

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