Question

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a speed of 0.75 m/s. The frictional force acting on the sled is one‑fifth of the combined weight of the child and the sled. If she travels for a distance of 25.5 m and her speed at the bottom is 4.30 m/s, calculate the angle that the slope makes with the horizontal.

Answer 1

θ = 13.7º

Step-by-step explanation:

• According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
• The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
• As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

`\Delta K + \Delta U = W_(nc) (1)`

• ΔK, is the change in kinetic energy, as follows:

`\Delta K = (1)/(2)* m* (v_(f) ^(2) - v_(0) ^(2)) (2)`

• ΔU, is the change in the gravitational potential energy.
• If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

`\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)`

• Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

`W_(nc) = F_(f) * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)`

• Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

`(1)/(2)* (v_(f) ^(2) - v_(0) ^(2)) = g*d* sin\theta -0.2*g*d`

• Replacing by the givens and the knowns, we can solve for sin θ, as follows:
  `(1)/(2)*( (4.30 m/s) ^(2) - (0.75 m/s)^(2)) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = (58.6 (m/s)2)/(250 (m/s)2) = 0.236`⇒ θ = sin⁻¹ (0.236) = 13.7º