Chapter 5, Problem 101CP

(a)

To determine

The acceleration of the block on sliding.

Answer to Problem 101CP

The acceleration of the block is $4.90\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the expression to calculate the acceleration of the block along an inclined plane.

  $a=g\sin \theta$

Here, a is the acceleration of the block, g is the acceleration due to gravity and $\theta$ is the angle of inclination.

Conclusion:

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for g $30.0°$ for $\theta$ in the above equation to calculate a.

  $\begin{array}{c}a=\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\sin 30.0°\\ =4.90\text{m}/{\text{s}}^{\text{2}}\end{array}$

Therefore, the acceleration of the block is $4.90\text{m}/{\text{s}}^{\text{2}}$.

(b)

To determine

The velocity of the block after it leaves the plane.

Answer to Problem 101CP

The velocity of the block is $3.13\text{m}/\text{s}$.

Explanation of Solution

Refer the equation (IV).

Write the expression to calculate the length of the inclined plane.

  $x=\frac{h}{\sin \theta }$                                                                                                                    (I)

Here, x is the length of inclined plane and h is the height of the inclined surface from the table.

Write the expression to calculate the velocity of the block.

  $v^2=u^2+2ax$                                                                                                          (II)

Here, v is the final velocity and u is the initial velocity.

Conclusion:

Substitute $0.500\text{m}$ for h and $30.0°$ for $\theta$ in the above equation (I) to calculate x.

  $\begin{array}{c}x=\frac{0.500\text{m}}{\sin 30.0°}\\ =1.00\text{m}\end{array}$

Substitute $4.90\text{m}/{\text{s}}^{\text{2}}$ for a, $1.00\text{m}$ for x and $0\text{m}/\text{s}$ for u in the above equation (II) to calculate v.

  $\begin{array}{c}{v}^2={\left(0\text{m}/\text{s}\right)}^2+2\left(4.90\text{m}/{\text{s}}^{\text{2}}\right)\left(1.00\text{m}\right)\\ v^2=9.80{\text{m}}^2/{\text{s}}^{\text{2}}\\ v=3.13\text{m}/\text{s}\end{array}$

Therefore, the velocity of the block is $3.13\text{m}/\text{s}$.

(c)

To determine

The distance from the table to the block when the later hits the floor.

Answer to Problem 101CP

The distance from the table is $1.35\text{m}$.

Explanation of Solution

Refer the equation (IV).

Write the expression to calculate the vertical displacement.

  $y=Vt+\frac{1}{2}g{t}^2$                                                                                                          (III)

Here, V is initial vertical velocity and t is the time taken to reach the ground.

Write the expression to calculate the vertical velocity of the block.

  $V=v\sin \theta$                                                                                                          (IV)

Use the equation (IV) to rewrite the equation (III).

  $y=\left(v\sin \theta \right)t+\frac{1}{2}g{t}^2$                                                                                               (V)

Write the expression to calculate the distance of the block from the table.

  $x=\left(v\cos \theta \right)t$                                                                                                        (VI)

Here, x is the distance from the table.

Conclusion:

Substitute $2.00\text{m}$ for y, $30.0°$ for $\theta$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for g and $3.13\text{m}/\text{s}$ for v in the above equation (V) to calculate t

  $\begin{array}{c}\left(3.13\text{m}/\text{s}\right)\left(\sin 30.0°\right)t+\frac{1}{2}\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)t^2=2.00\text{m}\\ \left(4.90\text{m}/{\text{s}}^{\text{2}}\right)t^2+\left(1.565\text{m}/\text{s}\right)t-2.00\text{m}=0\end{array}$

Solve the above quadratic equation to find t.

  $\begin{array}{c}t=\frac{-1.565\text{m}/\text{s}\pm \sqrt{{\left(1.565\text{m}/\text{s}\right)}^2-4\left(4.90\text{m}/{\text{s}}^{\text{2}}\right)\left(-2.00\text{m}\right)}}{2\left(4.90\text{m}/{\text{s}}^{\text{2}}\right)}\\ =\frac{-1.565\text{m}/\text{s}\pm 6.454\text{m}/\text{s}}{2\left(4.90\text{m}/{\text{s}}^{\text{2}}\right)}\\ =0.499\text{s}\end{array}$

Only positive root considered for the time t.

Substitute $30.0°$ for $\theta$ $0.499\text{s}$ for t and $3.13\text{m}/\text{s}$ for v in the above equation (VI) to calculate x.

  $\begin{array}{c}x=\left(3.13\text{m}/\text{s}\right)\left(\cos 30.0°\right)0.499\text{s}\\ =1.35\text{m}\end{array}$

Therefore, the distance from the table is $1.35\text{m}$.

(d)

To determine

The time interval between the releasing of the block and hitting the floor.

Answer to Problem 101CP

The time interval between the given events is $1.14\text{s}$.

Explanation of Solution

Writ the time required to reach the block to the bottom of the inclined plane.

  $t^{\prime }=\frac{v-u}{a}$

Here, $t^{\prime }$ is the time taken by the block to reach the bottom of the inclined plane.

Write the expression to calculate the time interval between the given events.

  $T=t+t^{\prime }$

Here, T is the time interval.

Rewrite the above equation using the expression for $t^{\prime }$.

  $T=t+\frac{v-u}{a}$

Conclusion:

Substitute $4.90\text{m}/{\text{s}}^{\text{2}}$ for a, $1.00\text{m}$ for x, $3.13\text{m}/\text{s}$ for v, $0.499\text{s}$ for t and $0\text{m}/\text{s}$ for u in the above equation to calculate T.

  $\begin{array}{c}T=0.499\text{s}+\frac{3.13\text{m}/\text{s}-0\text{m}/\text{s}}{4.90\text{m}/{\text{s}}^{\text{2}}}\\ =1.14\text{s}\end{array}$

Therefore, the time interval between the given events is $1.14\text{s}$.

(e)

To determine

Did the mass affect any of the calculation given above.

Answer to Problem 101CP

The mass of the block does not have any role on the calculation.

Explanation of Solution

In this case, the mass is moving along a frictionless inclined plane and therefore, there were no effect of mass for the mechanical calculations regarding the motion of the block.

The motion of the block is depends on the acceleration due to gravity as in the same case for an object under free fall.

Thus, the magnitude of mass of an object moving along a frictionless plane has no role in the mechanical calculations.

Conclusion:

Substitute $4.90\text{m}/{\text{s}}^{\text{2}}$ for a, $1.00\text{m}$ for x, $3.13\text{m}/\text{s}$ for v, $0.499\text{s}$ for t and $0\text{m}/\text{s}$ for u in the above equation to calculate T.

  $\begin{array}{c}T=0.499\text{s}+\frac{3.13\text{m}/\text{s}-0\text{m}/\text{s}}{4.90\text{m}/{\text{s}}^{\text{2}}}\\ =1.14\text{s}\end{array}$

Therefore, the mass of the block does not have any role on the calculation.