Question

In a circus performance, a monkey on a sled is given an initial speed of 4.3 m/s up a 24◦ incline. The combined mass of the monkey and the sled is 22.5 kg, and the coefficient of kinetic friction between the sled and the incline is 0.45. How far up the incline does the sled move?

Answer 1

Answer:

d = 1.15 m

Explanation:

• In absence of friction, the change in kinetic energy of the combined mass of the monkey and the sled, must be equal (with opposite sign), to the change in gravitational potential energy:

        ΔK = -ΔU

• When friction is not negligible, the change in mechanical energy, must be equal to the work done by non-conservative forces (kinetic friction in this case):

       ΔK + ΔU = Wnc (1)

• As the monkey + sled reach to the maximum distance up the incline, they will come momentarily to a stop, so the final kinetic energy is 0.

        $(K_{f} -K_{o}) = 0 - \frac{1}{2} * m*v_{o} ^{2} = -\frac{1}{2} *22.5kg*(4.3m/s)^{2} = -208.1J$

• The change in gravitational energy, can be written as follows:

        $(U_{f} - U_{o} ) = m*g*h - 0 = m*g*h = \\ \\ 22.5 kg*9.8 m/s2*d*sin (24 deg) = 89.7J*d$

• The sum of these two quantities, must be equal to the work done by the friction force, along the distance d up the incline:

        $W_{nc} = -\mu k*N*d$

• The normal force, always normal to the surface, must be equal and opposite to the component of the weight normal to the incline:

        $N = m*g*cos \theta = m*g*cos (24 deg) = \\ \\ 22.5 kg*9.8m/s2*0.913 = 201.4 N$

• Replacing in the equation for Wnc:

        $W_{nc} = -\mu k*N*d = -0.45*201.4 N*d = -90.6 N*d$

• We can return to the equation (1) and solve for d:

        $-208.1 J + 89.7N*d = -90.6N*d\\\\ d = \frac{208.1}{180.3} =1.15 m$