Answer to Question #137112 in Mechanics | Relativity for Subhasmita Dash

Question

A child of mass 20 kg slides down a 6.0 m long

slide, which makes an angle of 30° with the

horizontal. The coefficient of kinetic friction

between the child and the slide is 01. If the

child is at rest at the top of the slide, what is

her speed at the bottom ? How much energy is

transferred to thermal energy ?

Take g = 10 ms 2.

Answer 1

Explanations & Calculations

• When attempting these kinds of sums one should know how the components of the forces are written & the application of the conservation of energy under non conservative forces.
• Since no any motion perpendicular to the plane $\small R = mg\cos 30^0 = 100\sqrt3 N$
• Since the child moves under the friction. friction is at its highest : $\small f= \mu R = 10\sqrt3 N$
• Apply Newton's second law along the moving direction,

$\qquad\qquad \begin{aligned} \small mg\sin30-f &= \small ma\\ \small 200\times \frac{1}{2} - 10\sqrt3 &= \small 20a\\ \small a & =\small 4.13ms^{-2} \end{aligned}$

• Now the falling acceleration is known then apply $\small v^2 =u^2 +2as$ to calculate the velocity at the bottom.

$\qquad\qquad \begin{aligned} \small v^2 &= \small 0^2+2\times 4.13\times 6m\\ &= \small 49.56\,(ms)^2\\ \small v & = \small \bold{7.04ms^{-1}} \end{aligned}$

• Since the gravitational potential energy it has at the top exhaust to other possible forms as it reach down,

Potential energy = kinetic energy + work done against friction

$\qquad\qquad \begin{aligned} \small mgh &= \small \frac{1}{2}mv^2+fs\\ \small fs &= \small m(gh-\frac{1}{2}v^2)\\ &= \small 20\Big(10\times6\sin30-\frac{1}{2}\times49.56\Big)\\ &= \small \bold{104.4\,J} \end{aligned}$

• This work done against the friction is what transferred to heat.