Figure shows a block of mass m resting on a 20? slope. The block has coefficient

Question

Figure shows a block of mass m resting on a 20? slope. The block has coefficients of friction 0.77 and 0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have ?

Figure shows a block of mass m resting on a 20? slope. The block has coefficients of friction 0.77 and 0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg. If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have ?

Explanation / Answer

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.kg
as for the other equation change the mu_k(mg) to mu_k(mgcos20)
that should do

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