Use the equation for the acceleration
A = final velocity - initial velocity divided by time final - time initial
A= 54 - 32 / 8 - 0
A= 22 / 8
A= 2.75 m/s^2
Hope this helps!
Answer:
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Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Gravity lets all objects fall to the ground at the same speed, 9.8 m/s/s. If the force of gravity were stronger, such as 10 m/s/s, the rate of acceleration would be faster.
