Question

10A point object of mass 2.0 kg moves along x-axis with a velocity of + 4.0 m/s. A net horizontal force acting along the x-axis is applied to the object with the force-time profile shown. What is the total work that was done by the force?

• A. 20 J
• B. 25 J
• C. 65 J
• D. 81 J

Answer 1

Answer: (D)

81 J

Answer 2

To find the total work done by the force, we will use the Work-Energy Theorem, which states that the total work done on an object is equal to the change in its kinetic energy.

$W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

First, we need to determine the final velocity ($v_f$) of the object. We can do this using the Impulse-Momentum Theorem, which states that the impulse (J) applied to an object is equal to the change in its momentum.

$J = \Delta p = mv_f - mv_i$

The impulse (J) is the area under the Force-Time (F-t) graph. The given graph is a triangle. From the graph and problem description (referencing similar problem for missing values):

• The base of the triangle extends from $t = 0$ ms to $t = 6$ ms.
  Base ($B$) = $6 \text{ ms} = 6 \times 10^{-3} \text{ s}$
• The maximum height of the triangle (peak force) is not explicitly labeled in the provided figure for this question. However, a similar problem with identical text and options (except for the correct answer) specifies the peak force. Assuming this problem uses the same numerical values for the peak force:
  Height ($H$) = $4.0 \text{ kN} = 4.0 \times 10^3 \text{ N}$

1. Calculate the Impulse (J):
   The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
   $J = \frac{1}{2} \times B \times H$
   $J = \frac{1}{2} \times (6 \times 10^{-3} \text{ s}) \times (4.0 \times 10^3 \text{ N})$
   $J = \frac{1}{2} \times 24 \text{ Ns}$
   $J = 12 \text{ Ns}$

2. Calculate the Final Velocity ($v_f$):
   Given mass ($m$) = $2.0 \text{ kg}$
   Initial velocity ($v_i$) = $+4.0 \text{ m/s}$
   Using the Impulse-Momentum Theorem:
   $J = mv_f - mv_i$
   $12 \text{ Ns} = (2.0 \text{ kg}) v_f - (2.0 \text{ kg})(4.0 \text{ m/s})$
   $12 = 2v_f - 8$
   $2v_f = 12 + 8$
   $2v_f = 20$
   $v_f = 10 \text{ m/s}$

3. Calculate the Total Work Done (W):
   Using the Work-Energy Theorem:
   $W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
   $W = \frac{1}{2}(2.0 \text{ kg})(10 \text{ m/s})^2 - \frac{1}{2}(2.0 \text{ kg})(4.0 \text{ m/s})^2$
   $W = (1.0)(100) - (1.0)(16)$
   $W = 100 - 16$
   $W = 84 \text{ J}$

The calculated total work done by the force is 84 J. Comparing this with the given options: (A) 20 J (B) 25 J (C) 65 J (D) 81 J

Since 84 J is not among the options, we choose the closest option, which is 81 J. This suggests a slight numerical discrepancy in the problem's intended values (e.g., if the peak force was 3.9 kN, the work done would be approximately 81 J).