Q106P - University Physics with Modern Physics

Question

A 70-kg person rides in a 30-kg cart moving at 12 m/s at the top of a hill that is in the shape of an arc of a circle with a radius of 40 m. (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart can travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

Step-by-Step Solution

Answer

(a) The apparent weight of the person is 434 N.

(b) The maximum velocity is 19.8 m/s, and it does not depend on the mass of the cart or person.

Step 1: Describe the Newton’s second law and centripetal acceleration

According to Newton’s second law, the linear force is given by,

F = ma

Here, F is linear force, m is the mass of object, and a is acceleration of object.

 

The centripetal acceleration is given by,

${\mathbf{a}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{r}}$ 

Here, v is velocity, and r is radius of curvature.

Step 2: Determine the apparent weight of the person

(a)

 

Draw the free-body diagram of the give situation as follows.

According to the Newton’s second law, the centripetal force is given by,

$$\begin{gathered} \sum _{ }{F}_y=-ma \\ N-mg=-m{a}_c \\ N-mg=-\frac{m{v}^2}{r} \\ N=mg-\frac{m{v}^2}{r} .......\left(1\right) \end{gathered}$$ 

 

Substitute 70 kg for m, $9.8m/s^2$ for g, 12 m/s for v, and 40 m for r in equation (1).

$$\begin{gathered} \mathrm{N}=\left(70 \mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\frac{\left(70 \mathrm{kg}\right){\left(12 \mathrm{m}/\mathrm{s}\right)}^2}{40 \mathrm{m}} \\ =686-252 \\ =434 \mathrm{N} \end{gathered}$$ 

 

Therefore, the apparent weight of the person is 434 N.

Step 3: Determine the maximum speed that the cart and determine whether the answer depend on the mass of the cart or the mass of the person

(b)

 

The speed at which the cart will slide off means that N=0. From equation (1),

$$\begin{gathered} 0=\mathrm{mg}-\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \mathrm{mg}=\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \mathrm{g}=\frac{{\mathrm{v}}^2}{\mathrm{r}} \\ {\mathrm{v}}_{\max }=\sqrt{\mathrm{gr}} ..........\left(2\right) \end{gathered}$$

 

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2$  for g, and 40 m for r in equation (2).

$$\begin{gathered} {\mathrm{v}}_{\max }=\sqrt{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(40 \mathrm{m}\right)} \\ =19.8 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Therefore, the maximum velocity is 19.8 m/s.

 

From the equation (2), it can be observed that the maximum velocity does not depend on the mass of the cart or person.