Question

A four-person bobsled$\left(\mathrm{total}\mathrm{mass}=630\mathrm{kg}\right)$comes down a straightaway at the start of a bobsled run. The straightaway is 80.0 mlong and is inclined at a constant angle of$\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{°}$with the horizontal. Assume that the combined effects of friction and air drag produce on the bobsled a constant force of 62.0 Nthat acts parallel to the incline and up the incline. Answer the following questions to three significant digits.

(a) If the speed of the bobsled at the start of the run is 6.20 m/s, how long does the bobsled take to come down the straightaway?

(b) Suppose the crew is able to reduce the effects of friction and air drag to 42.0 N. For the same initial velocity, how long does the bobsled now take to come down the straightaway?

Short answer

$$\begin{gathered} \mathbf{a}\mathbf{)}\mathbf{}6.80\mathrm{s} \\ \mathbf{b}\mathbf{)}\mathbf{}6.76\mathrm{s} \end{gathered}$$

Step-by-step solution

Given data

$$\begin{gathered} \mathrm{Total}\mathrm{mass}\left(m\right)=630\mathrm{kg} \\ \mathrm{Lenght}\left(s\right)=80.0\mathrm{m}\mathrm{and}\mathrm{is}\mathrm{inclined}\mathrm{at}\mathrm{a}\mathrm{constant}\mathrm{angle}\mathrm{of}10.2°\mathrm{with}\mathrm{the}\mathrm{horizontal}. \\ \mathrm{Force}\left(f\right)=62.0\mathrm{N} \\ \mathrm{Initial}\mathrm{velocity}\mathrm{is},u=6.20\mathrm{m}/\mathrm{s} \end{gathered}$$

Understanding the concept

The problem deals with Newton’s second law of motion. It tells us that the acceleration a of an object is affected by the net force F acting upon the object and the mass of m the object.

Formula:

F = ma

a)Calculate how long the bobsled takes to come down the straightaway if the speed of the bobsled at the start of the run is 6.20 m/s

The component of the weight along the incline (with downhill understood as the positive direction) is,

$mg\sin \theta$

Where m is the mass and is$\theta$ the angle.

We can write the balance force equation as,

$mg\sin \theta -f=ma$

Substitute the values in the above equation, and we get,

$$\begin{gathered} 630\times 9.8\times \sin 10.2°-62=630\times a \\ a=1.937\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{a}=1.64\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

We know that from the equation of motion, we can write the expression for distance as,

$s=ut+1/2a{t}^2$

Substitute the values in the above equation, and we get,

$80.0\mathrm{m}=\left(6.20\mathrm{m}/\mathrm{s}\right)\mathrm{t}+\left(1/2\right)\left(1.64\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2$

The above equation is a quadratic equation of t; we can find the root as,

t = 6.80 s

Thus, the bobsled takes 6.80 s to come down straightaway.

b) Calculate how long the bobsled now takes to come down the straightaway for the same initial velocity if the effects of friction and air drag is reduced to 42.0 N

$\mathrm{Force}\left(\mathrm{f}\right)=42.0\mathrm{N}$

We can write the balance force equation as,

$mg\sin \theta -f=ma$

Substitute the values in the above equation, and we get,

$$\begin{gathered} 630\times 9.8\times \sin 10.2°-42=630\times a \\ a=1.66\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

We know that from the equation of motion, we can write the expression for distance as,

$s=ut+1/2a{t}^2$

Substitute the values in the above equation, and we get,

$80.0\mathrm{m}=\left(6.20\mathrm{m}/\mathrm{s}\right)\mathrm{t}+\left(1/2\right)\left(1.66\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2$

The above equation is a quadratic equation of t; we can find the root as,

t = 6.76 s

Thus, the bobsled takes 6.76 s to come down the straightaway.