Question

Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a
rope that is angled 30o above the floor. The tension is a constant 30 N and the coefficient of
friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m

Answer 1

Answer on Question #45366, Physics, Mechanics | Kinematics | Dynamics

Question:

Susan's 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30o above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul's speed after being pulled 3.0 m

Answer:

$F_{fr}$ - friction force

$F$ - pulling force

The law of conservation of energy:

$$\Delta E = W$$

where $\Delta E$ - change of body's energy, $W$ - work of all forces acting on the body

Work can be expressed by the following equation:

$$W = F d \cos \theta$$

where $F$ is the force, $d$ is the displacement, and the angle $\theta$ is defined as the angle between the force and the displacement vector.

Work of the friction force equals:

$$W_{fr} = F_{fr} d \cos 180{}^\circ$$

Work of the pulling force equals:

$$W_p = F d \cos 30{}^\circ$$

change of body's energy equals:

$$\Delta E = \frac{m v^2}{2}$$

Newton's first law of motion on y-axis:

$$F \sin 30{}^\circ + N = m g$$

Therefore, normal force equals:

$$N = m g - F \sin 30{}^\circ$$

Force of friction equals:

$$F_{fr} = \mu N = \mu (m g - F \sin 30{}^\circ)$$

Therefore:

$$\frac{m v^2}{2} = F d \cos 30{}^\circ + \mu (m g - F \sin 30{}^\circ) d \cos 180{}^\circ = \frac{\sqrt{3}}{2} F d - \mu \left(m g - \frac{F}{2}\right) d$$ $$v = \sqrt{\sqrt{3} \frac{F}{m} d - 2 \mu g d + \frac{F}{m} d} = 2.37 \frac{m}{s}$$

Answer: $2.37 \frac{m}{s}$

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