QuestionJune 17, 2025

A mover is applying a force on a crate up an incline.The incline is slightly rough with a coefficient of kinetic friction of 0.150 If the piano is not accelerating, that is moving up the incline with constant speed, what is the magnitude of the work done by the mover if the incline is 3.00 m long and the piano has a mass of 180 kg. The angle of the incline is 10.0^circ Use three significant digits please. square disappointed

A mover is applying a force on a crate up an incline.The incline is slightly rough with a coefficient of kinetic friction of 0.150 If the piano is not accelerating, that is moving up the incline with constant speed, what is the magnitude of the work done by the mover if the incline is 3.00 m long and the piano has a mass of 180 kg. The angle of the incline is 10.0^circ Use three significant digits please. square disappointed
A mover is applying a force on a crate up an incline.The incline is slightly rough with a coefficient of kinetic friction of 0.150 If the piano is not accelerating, that is moving up the incline with constant speed, what is the magnitude of the work done by the mover if the incline is 3.00 m long and the piano has a mass of 180 kg. The angle of the incline is 10.0^circ Use three significant digits please. square disappointed

Solution
4.1(235 votes)

Answer

5,130 \, \text{J} Explanation 1. Calculate the gravitational force component along the incline The gravitational force component along the incline is F_{\text{gravity}} = mg \sin(\theta), where m = 180 \, \text{kg}, g = 9.81 \, \text{m/s}^2, and \theta = 10.0^\circ. So, F_{\text{gravity}} = 180 \times 9.81 \times \sin(10.0^\circ). 2. Calculate the frictional force The normal force F_{\text{normal}} = mg \cos(\theta). The frictional force F_{\text{friction}} = \mu_k F_{\text{normal}}, where \mu_k = 0.150. So, F_{\text{friction}} = 0.150 \times 180 \times 9.81 \times \cos(10.0^\circ). 3. Calculate the total force exerted by the mover Since the piano moves with constant speed, the mover's force F_{\text{mover}} must balance both gravity and friction: F_{\text{mover}} = F_{\text{gravity}} + F_{\text{friction}}. 4. Calculate the work done by the mover Work done W = F_{\text{mover}} \times d, where d = 3.00 \, \text{m}.

Explanation

1. Calculate the gravitational force component along the incline<br /> The gravitational force component along the incline is $F_{\text{gravity}} = mg \sin(\theta)$, where $m = 180 \, \text{kg}$, $g = 9.81 \, \text{m/s}^2$, and $\theta = 10.0^\circ$. So, $F_{\text{gravity}} = 180 \times 9.81 \times \sin(10.0^\circ)$.<br /><br />2. Calculate the frictional force<br /> The normal force $F_{\text{normal}} = mg \cos(\theta)$. The frictional force $F_{\text{friction}} = \mu_k F_{\text{normal}}$, where $\mu_k = 0.150$. So, $F_{\text{friction}} = 0.150 \times 180 \times 9.81 \times \cos(10.0^\circ)$.<br /><br />3. Calculate the total force exerted by the mover<br /> Since the piano moves with constant speed, the mover's force $F_{\text{mover}}$ must balance both gravity and friction: $F_{\text{mover}} = F_{\text{gravity}} + F_{\text{friction}}$.<br /><br />4. Calculate the work done by the mover<br /> Work done $W = F_{\text{mover}} \times d$, where $d = 3.00 \, \text{m}$.
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