A 58-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to t

Question

A 58-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through a distance of 1.0 m before leaving the skater's hand at a speed of 6.0 m/s. Assume the skater and the ball to be point-like, the surface to be frictionless, and ignore air resistance. Use a vertical y-axis with the positive direction pointing up and a horizontal x-axis with the positive direction pointing to the right. Determine the acceleration of the ball during the throw. Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the acceleration is to the right and negative value if the direction of the acceleration is to the left. Determine the acceleration of the skater during the throw. Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the acceleration is to the right and negative value if the direction of the acceleration is to the left.

Explanation / Answer

here given are

vf=6m/sec

m1=58kg

m2=5kg

d=1m

vf^2-vi^2=2ad

now interchanging the equation we get
a= (vf^2-vi^2)/2d
=(6.0^2 - 0^2/)2(1m)
a=18m/s^2

force on the 5kg medicine ball

=ma
=(5kg)(18m/s^2)
=90N
a = F/m
a=98/58kg
a=1.68m/s^2

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