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A child and sled with a combined mass of 48.6 kg slide down a frictionless hill that is 8.02 m high. If the sled starts from rest, what is its speed at the bottom of the hill?

and

A person doing a chin-up weighs 755 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exerts an upward force of 381 N on the torso. If the upward movement starts from rest, what is the person's speed at this point?

2007-12-17 12:39:14 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

v=sqrt(2*9.8*8.02)
12.54 m/s

This is derived from the energy equations
.5*m*v^2=m*g*h
solving for v
v=sqrt(2*g*h)

Chin up:

using f=m*a
381*2-755=755*a/9.81
solve for a
a=0.091 m/s^2

the time it takes to ascend 25 cm
25=.5*a*t^2
solve for t
2.34 seconds
to find speed
v=a*t
0.213 m/s



j

2007-12-19 08:41:48 · answer #1 · answered by odu83 7 · 0 0

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