A 2.200-kg car moving down a road with a slope (grade) of 14% at a constant spee

Question

A 2.200-kg car moving down a road with a slope (grade) of 14% at a constant speed of 11 m/s. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)? A 2.319-kg car is moving down a road with a slope (grade) of 11% while slowing down at a rate of 3.8 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)? A 1.780-kg car a moving down a road with a slope (grade) of 39% while speeding up at a rate of 1 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)? A 1.771-kg car is moving down a road with a slope (grade) of 11% while speeding up at a rate of 3.1 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)? A 2.073-kg car ts moving up a road with a slope (grade) of 32% at a constant speed of 14 m/s What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)? A 2.199-kg car is moving up a road with a slope (grade) of 29% while slowing down at a rate of 2.1 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)? A 1.315-kg car is moving up a road with a slope (grade) of 15% while slowing down at a rate of 4.5 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)? A 1.521-kg car is moving up a road with a slope (grade) of 27% while speeding up at a rate of 1.7 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i e . down the slope)?

Explanation / Answer

14% slope = inverse tan ( 14 / 100 ) = 7.9796 °

m = car mass = 2,200 kg

If the car is at constant speed, then driving and friction forces are equal.

the friction force act opposite to the direction of motion.

The driving force (and therefore the friction forces) = m * g * sine 7.9796 ° = 2,989.2475 N

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