Figure shows a block of mass m resting on a 20 degree slope. It is connected via

Question

Figure shows a block of mass m resting on a 20 degree slope. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0kg .

Given:

m2= 2.0kg

uk= .78

us= .32

height btw m2 and the ground= 2.0m

height btw m1 and the ground= 2.5m

Find (PLEASE ANSWER 1-6)

1) what direction will they travel and why?

2) what is the acceleration (a)? What are velocities (v)?

3) how fast is block 2 moving when it hits the ground?

4) how far does (m)1 travel? (remember once (m2) hits the ground, (m1) still travels a certain distance before it falls back and the rope has Tension in it again).

5) what is the tension (T) after m1 falls back down?

6) how long does it take for (m1) to return to the resting position on the slope? thankyou for looking. I really appreciate the help.

PLEASE ANSWER 1-6!!!!!!!!!!!!!

Explanation / Answer

please upload the fig......

i just assumed som e things.hope it may help

the force of friction is equal to mu(m)(g)
the pullling force is 2(g)
so set 2(g)= mu_s(m)(g)
2/mu_s=m
(2g)-mu_k(mg)=f_net
F_net/m (of the block)= acceleration
yes im sorry the equation is incorrect. the force of friction is equal to mu(normal force) the normal force in this case is perpendicular to the plane making it = mgcos20
in this case the pulling force is still 2g so the new equation is 2g=mu_s(mgcos20)
2g/(mu_s(g)cos20)=m
m=2.65kg
as for the other equation change the mu_k(mg) to mu_k(mgcos20)

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