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A 52.7 W motor applies a force to a 99.7 kg object for 54.5 seconds over a distance of 3.97 meters. At what rate will the object accelerate?

Answer :

The motor will accelerate the object at a rate of approximately[tex]3.72 m/s^2.[/tex]

Given:

- Force (F) = 52.7 W

- Mass (m) = 99.7 kg

- Time (t) = 54.5 s

- Distance (d) = 3.97 m

1: Calculate Work Done (W):

[tex]\[W = F \times d\][/tex]

[tex]\[W = 52.7 \, \text{W} \times 3.97 \, \text{m} = 208.819 \, \text{J}\][/tex]

2: Calculate Acceleration (a) using Work-Energy Principle:

[tex]\[W = \frac{1}{2} \times m \times v^2\][/tex]

[tex]\[208.819 \, \text{J} = \frac{1}{2} \times 99.7 \, \text{kg} \times v^2\][/tex]

3: Solve for velocity (v):

[tex]\[v^2 = \frac{2 \times 208.819}{99.7}\][/tex]

[tex]\[v^2 = 4.183\][/tex]

[tex]\[v = \sqrt{4.183} = 2.045 \, \text{m/s}\][/tex]

4: Calculate Acceleration (a) using kinematic equation:

[tex]\[a = \frac{v_f - v_i}{t}\][/tex]

[tex]\[a = \frac{2.045 \, \text{m/s} - 0}{54.5 \, \text{s}}\][/tex]

[tex]\[a \approx 0.0374 \, \text{m/s}^2\][/tex]

Therefore, the motor will accelerate the object at a rate of approximately [tex]3.72 m/s^2.[/tex]

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