A 5kg block starts at a height of 10m above the ground and slides 20m down a rou

Question

A 5kg block starts at a height of 10m above the ground and slides 20m down a rough surfaced incline. The work done by the frictional force in sliding down the incline is 240 J. It reaches the bottom of the incline and then enters a frictionless vertical loop of radius 2m. The speed of the block at the top of the hoop is 4.65m/s. The block performs one revolution and exits the loop.

1)Find the normal force at the top of the circular hoop

2)Find the angular speed at the top of the hoop

3)Use energy methods to find the speed at the bottom of the hoop..

4) Please give details answer to this problem Dont Skip any step in order to undertand what you did. Thanks a lot in advance.

Explanation / Answer

given

m = 5 kg

R = 2 m

1) Let N is the normal force on the block at the top of the loop.

net force acting on block, Fnet = N + m*g

m*a_rad = N + m*g

m*v^2/R = N + m*g

==> N = m*v^2/R - m*g

= 5*4.65^2/2 - 5*9.8

= 5.056 N

2) angular speed, w = v/R

= 4.65/2

= 2.325 rad/s

3) mechaincal energy at the top of the loop = mechanical energy at the bottom of loop

0.5*m*v_top^2 + m*g*h = 0.5*m*v_bottom^2

0.5*m*v_top^2 + m*g*(2*R) = 0.5*m*v_bottom^2

v_top^2 + 2*g*(2*R) = v_bottom^2

==> V_bottom = sqrt(v_top^2 + 4*g*R)

= sqrt(4.65^2 + 4*9.8*2)

= 10 m/s

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