Recent content by Creepypunguy

So by solving equations 1 and 3 I got mglsinθ+mgxsinθ = (1/2)kx2 0 = (k/2)x2 - mgsinθx - mglsinθ quadratic formula x = (mgsinθ ± √[mgsinθ(mgsinθ+2kl)] ) / k

Thanks guys!

After the above corrections I came to a final answer of: μk = (1/2)(μs - tanθ)

Homework Statement A block of mass m starts from rest at a height h and slides down a frictionless plane inclined at angle θ with the horizontal, as shown below. The block strikes a spring of force constant k. Find the distance the spring is compressed when the block momentarily stops...

Ok so starting now from (1/2)kx2 = mgx(sinθ+μkcosθ) Sub in x from previous solution; (-1/2)mg(μscosθ+sinθ) = mg(sinθ+μkcosθ) With a small amount of arithmetic becomes; μk= (-1/2)(μs+3tanθ) Does that final solution check out?

So the Energy at the initial point would just be the potential energy of the spring?

Homework Statement http://imgur.com/xIckJqW A block of mass m rests on a plane inclined at an angle θ with the horizontal. A spring with force constant k is attached to the block. The coefficient of static friction between the block and plane is μs. The spring is pulled upward along the...

Homework Statement A quarry crane is used to lift massive rocks from a quarry pit. Consider the simplified model of such a crane shown in the figure. (Figure 1) The ends of two poles are anchored to the ground at the same point (point O). From this point, one pole rises vertically and the...