Question

A block of mass 5Kg is released from point A and slides on the frictionless track shown in Figure P8.6. Determine (a) the block’s speed at points B and C and (b) the net work done by the gravitational force on the block as it moves from point A to point C.

Short answer

(a) The block's speed at points B and C are 5.94 and 7.67 m/s respectively.

(b) The net work done by the gravitational force on the block as it moves from point A to point C is 147.1 J

Step-by-step solution

Given data:

Mass of the block, m=5kg

Height of point A, h_a =5m

Height of point B, h_b =3.2 m

Height of point C, h_c=2m

Step 2:Potential and kinetic energy

The potential energy of a mass at a height from the ground is,

p=mgh.......................................................1

The kinetic energy of a mass m moving with a velocity v is

$K=\frac{1}{2}m{v}^2$...................................................2

Step 3:(a) Determining the velocities of the block at points B and C:

From equation (I), the potential energy of the block at point A is

The block is released from rest at point A, hence it has no kinetic energy at point A.

Thus, the total energy of the block is

$$\begin{gathered} E=P_A \\ =245\text{J} \end{gathered}$$

The total energy of a body is conserved. Let the velocity of the block at point B be V_B . Thus, from equations (I) and (II)

Substitute the values to get

Step 4:(b) Determining the work done to move the block from points A to C:

According to the work-energy theorem, work done is equal to the change in kinetic energy. Kinetic energy at point A is zero. From equation (II), kinetic energy at point C is

$$\begin{gathered} K_C=\frac{1}{2}m{v}_C^2 \\ =\frac{1}{2}\times 5\text{kg}\times {\left(7.67\text{m/s}\right)}^2 \\ =147.1·\left(1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right) \\ =147.1\text{J} \end{gathered}$$

This is also equal to the work done on the block.