A .50kg mass begins at rest and slides down a plane tilted 25 degrees above the

Question

A .50kg mass begins at rest and slides down a plane tilted 25 degrees above the horizontal.

The coef. of friction between the inclined plane and the mass is .20.  

1) How much time will it take for the mass to slide, from rest, to the bottom of the 1.5 m inlcined plane?

2) What is the work done on the mass by gravity.

3) What is the work done by friction?

4) At the bottom of the inclined plane, the mass slides along a smooth horizontal surface and compresss a spring of force constant 500 N/m.  What is the distance the spring is compressed?

Explanation / Answer

m=0.5kg,theta=25 degrees,mu=0.2

accelarion of mass along incline=(g*sin25-mu*g*cos25)=2.365.

1)So,time taken=sqrt(2*1.5/2.365)=1.126s.

2)work done=mglsin25=0.5*9.8*1.5*sin25=3.106 Nm

3)work done=-mu*m*g*cos25*l=-1.3322N/m

4)Work done by (gravity+friction+spring)=0

work done by spring=1.3322-3.106=-1.7738Nm

1/2*k*x^2=1.7738

x=0.084m

=84cm

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