Problem

The 50-kg crate of Prob. 3/1 is now projected down an incline as shown with an initial s...

The 50-kg crate of Prob. 3/1 is now projected down an incline as shown with an initial speed of 7 m/s. Investigate the time t required for the crate to come to rest and the corresponding distance x traveled if (a) θ = 150 and (b) θ = 300.

Step-by-Step Solution

Solution 1

Balancing the forces in the perpendicular direction of motion (y-Coordinate direction), it is obtained:

$$ N=m g \cos \theta $$

Again balancing the forces in the direction of motion (x-Coordinate direction), it is obtained:

$$ \begin{aligned} &m a=\mu_{k} N-m g \sin \theta \\ &m a=\mu_{k} m g \cos \theta-m g \sin \theta \\ &a=\mu_{k} g \cos \theta-g \sin \theta \end{aligned} $$

Here \(a\) is the acceleration, \(g\) is acceleration due to gravity, \(\mu_{k}\) is the coefficient of kinetic friction and \(\theta\) is inclination angle.

Substitute \(9.81 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 0.4\) for \(\mu_{k}\) and \(15^{\circ}\) for \(\theta=15^{\circ}\) in above relation.

$$ \begin{aligned} a &=9.81 \times\left(0.4 \times \cos 15^{\circ}-\sin 15^{\circ}\right) \\ &=1.2513 \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$

Calculate the distance travelled by using kinematic equation.

$$ \begin{aligned} &v_{0}^{2}=2 a x \\ &x=\frac{v_{0}^{2}}{2 a} \end{aligned} $$

Here \(v_{0}\) is the initial speed and \(a\) is the acceleration.

Substitute \(7 \mathrm{~m} / \mathrm{s}\) for \(v_{0}\) and \(1.2513 \mathrm{~m} / \mathrm{s}^{2}\) for \(a\) in above equation.

$$ \begin{aligned} x &=\frac{7^{2}}{2 \times 1.2513} \\ &=19.5799 \mathrm{~m} \end{aligned} $$

Therefore, distance travelled \(x\) is \(19.6 \mathrm{~m}\).

Calculate time required for the crate to come to rest.

$$ \begin{aligned} &v_{0}=a t \\ &t=\frac{v_{0}}{a} \end{aligned} $$

Substitute \(7 \mathrm{~m} / \mathrm{s}\) for \(v_{0}\) and \(1.2513 \mathrm{~m} / \mathrm{s}^{2}\) for \(a\) in above equation.

$$ \begin{aligned} t &=\frac{7}{1.2513} \\ &=5.594 \mathrm{~s} \end{aligned} $$

Therefore, time required for the crate to come to rest is \(5.59 \mathrm{~s}\).

(b) Again, for \(\theta=30^{\circ}\), it is obtained that:

$$ \begin{aligned} a &=\mu_{k} g \cos \theta-g \sin \theta \\ &=9.81 \times\left(0.4 \times \cos 30^{\circ}-\sin 30^{\circ}\right) \\ &=-1.51 \frac{\mathrm{m}}{\mathrm{s}^{2}} \end{aligned} $$

As this acceleration is negative, the crate will move downwards and will not eventually come to rest.

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