To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch
1 answer:
Answer:
man upward acceleration is 0.14m/s^2
Explanation:
given data:
mass of man = 72 kg
downward force = 360 N
The mass of man of weight 72 kg is hang from two sections of rope, one section pf rope ties around man waist and other section is ties in man hands. when he pulls down the rope with 360 N force then each section of rope pulls with 360 N
we know that
Weight= mass × gravity= 72kg × 9.8 = 705.6N
Force = mass× acceleration
Force= -705.6 + (2 × 358) = 10.4 N
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Answer:
(a) Height is 4.47 m
(b) Height is 4.37 m
Solution:
As per the question:
Initial velocity of teh ball,
Angle made by the ramp,
Distance traveled by the ball on the ramp, d = 5.00 m
Now,
(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:
where
H =
g =
= 19.06 m/s
Now, maximum height attained is given by:
Height from the ground =
(b) now, considering the coefficient of friction bhetween ramp and the ball, :
velocity can be given by the eqn-3 of motion:
= 18.7 m/s
Now, maximum height attained is given by:
Height from the ground =