You drive your 1,266-kg car at a constant speed of 20 mph down a ramp that has a grade of 12 %. Use conservation of energy to find the work done by the frictional force acting from the road on the tires. Define positive direction in the forward direction.
Someone please help me :(
Someone please help me :(
-
+ direction = down the ramp.
Assume grade of 12% means 12% of 90 deg = 10.8 deg above horizontal.
The acceleration is 0 so the magnitude of the frictional force up the ramp must equal the magnitude of the gravity force down the ramp which is:
f = 1226*g*sin(10.8) N = 2250 N
Now we only have a speed, but no time or distance specified so energy or Work Done can not be calculated. If you knew the distance d (in m) travelled down the ramp the WD by friction would be:
WD = -2250 * d Joule
Assume grade of 12% means 12% of 90 deg = 10.8 deg above horizontal.
The acceleration is 0 so the magnitude of the frictional force up the ramp must equal the magnitude of the gravity force down the ramp which is:
f = 1226*g*sin(10.8) N = 2250 N
Now we only have a speed, but no time or distance specified so energy or Work Done can not be calculated. If you knew the distance d (in m) travelled down the ramp the WD by friction would be:
WD = -2250 * d Joule