Answer :
Answer:
0.394
Explanation:
Complete question:
Paul is pushing the couch up the ramp into the moving truck at a constant velocity. If he pushes with a force of 241 N directly to the right (NOT AT ANY ANGLE), what is the coefficient of kinetic friction? The mass of the couch is 36.3 kg, and the angle of the ramp is 21.5°.
Since the velocity is constant, the acceleration is zero.
Take the sum of the force along the plane according to Newton's second law:
[tex]\sum Fx = ma_x\\Since \ a = 0m/s^2\\\sum Fx = 0\\Fm-Ff = 0\\Fm = Ff\\mgsin\theta = \mu mgcos \theta\\sin\theta = \mu cos\theta\\\mu = \frac{sin\theta}{cos\theta}\\ \mu = tan \theta\\[/tex]
Given
[tex]\theta = 21.5^0\\[/tex]
[tex]\mu = tan 21.5^0\\\mu = 0.394\\[/tex]
Hence the coefficient of kinetic friction 0.394
The coefficient of kinetic friction between the couch and the ramp is 0.68.
The given parameters;
- mass of the couch, m = 36.3 kg.
The net force on the couch is calculated as follows;
[tex]F - \mu_k F_n = ma[/tex]
where;
- [tex]\mu_k[/tex] is the coefficient of kinetic friction
- a is the acceleration of the couch
At constant velocity, the acceleration of the couch = 0
[tex]F - \mu_k F_n = 0\\\\\mu_kF_n = F\\\\\mu_k = \frac{F}{F_n} \\\\\mu_k = \frac{F}{mg} \\\\\mu_k = \frac{241}{36.3 \times 9.8} \\\\\mu_k = 0.68[/tex]
Thus, the coefficient of kinetic friction between the couch and the ramp is 0.68.
"Your question is not complete, it seems to be missing the following information;"
Paul is pushing the couch up the ramp into the moving truck at a CONSTANT VELOCITY. If he pushes with a force of 241 N directly to the right (NOT AT ANY ANGLE), what is the coefficient of kinetic friction? The mass of the couch is 36.3 kg.
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