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Question
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- a (4.2 \times 103 - \text{n}) piano is wheeled up a (3.5 -\text{m}) ramp at a constant speed. the ramp makes an angle of (30.0^circ) with the horizontal. find the work done by a man wheeling the piano up the ramp.
Step1: Recall the work formula for force and displacement
The work done \( W \) by a force \( F \) over a displacement \( d \) when the force is at an angle \( \theta \) to the displacement is given by \( W = Fd\cos\theta \). But in this case, the force applied to move the piano at constant speed up the ramp should balance the component of the weight along the ramp. The weight of the piano is \( F_g=4.2\times 10^{3}\ N \), and the component of the weight along the ramp is \( F_g\sin\theta \) (since the ramp angle is \( \theta = 30.0^\circ \) with the horizontal, the component along the ramp is opposite to the angle in the right - triangle formed by the weight, normal force and the ramp). However, when moving at constant speed, the applied force \( F \) is equal to the component of the weight along the ramp, i.e., \( F = F_g\sin\theta \). The displacement \( d \) along the ramp is \( 3.5\ m \). But we can also think of the work done against gravity (since at constant speed, the net work is zero, and the work done by the man is equal to the work done against gravity). The work done against gravity is \( W = mgh \), but we can also express \( h \) (the vertical height) as \( d\sin\theta \), so \( W=F_g\times h=F_g\times d\sin\theta \).
Step2: Substitute the values
We know that \( F_g = 4.2\times 10^{3}\ N \), \( d = 3.5\ m \), and \( \theta=30.0^\circ \). We know that \( \sin(30.0^\circ)=\frac{1}{2} = 0.5 \).
So, \( W=F_g\times d\times\sin\theta=(4.2\times 10^{3}\ N)\times(3.5\ m)\times\sin(30.0^\circ) \)
First, calculate \( (4.2\times 10^{3})\times3.5 = 4.2\times3.5\times 10^{3}=14.7\times 10^{3} \)
Then, multiply by \( \sin(30.0^\circ) \): \( 14.7\times 10^{3}\times0.5 = 7.35\times 10^{3}\ J \)
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The work done by the man is \( 7.35\times 10^{3}\ J \) (or \( 7350\ J \))