Based off the image, m=25kg L=10m musubk=.1 and theta=30 degrees. What would a f

Question

Based off the image, m=25kg L=10m musubk=.1 and theta=30 degrees. What would a force diagram look like? what is the magnitude and direction of the normal force? what is the magnitude and direction of the friction force? what is the magnitude and riction of the acceleration of the object? if the mass is initally at the orgin on the slope and the velocity of 3m/s what is the coordinate of the mass at the time showiwn in the picture? if it goes all the way down the hill, at what time does it reach the bottom ofthe hill? if it does come to a stop before that time, when and where does it come to a stop? show all work if possible.

Explanation / Answer

I understand the direction of friction is to the left(-) but Im not sure how to calculate the magnitude. For this problem the magnitude is 918 N. If you can provide the steps in how they got to 918 N that would be great The friction force causes the 10.0 gram bullet to decelerate from 205 m/s to 0 m/s as it travels 22.9 cm = 0.229 m The equation below is very useful, when you do not know the time required to decelerate!! (Final velocity)^2 – (Initial velocity)^2 = 2 * acceleration * distance Final velocity = 0 m/s Initial velocity = 205 m/s Distance = 0.229 m 0^2 – 205^2 = 2 * acceleration * 0.229 Acceleration = -205^2 ÷ 0.458 Force = mass * acceleration Mass = 10.0 g = 0.010 kg Acceleration = -205^2 ÷ 0.458 Friction force = 0.010 * (-205^2 ÷ 0.458) Friction force = -917.6 N The negative sign means the friction force is in the direction opposite to the motion of the bullet. Friction force = 917.6 N to the LEFT

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