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

as all the questions are of same type:

let mass of the car=m kg.

slope be x.

angle of the slope=theta=arctan(x)

let acceleration in positive direction is a m/s^2.

component of weight along down the slope=m*g*sin(theta)

component of weight perpendicular to the slope=m*g*cos(theta)

friction force will be opposite to the direction of motion.

hence as in each case the car is moving down the slope, friction force will be in the upward direction

along the slope.

let friction force magnitude be F.

then writing force balance equation in down the slope direction:

m*g*sin(theta)-F=m*a

==>F=m*(g*sin(theta)-a)

part a:

x=0.13

==>theta=7.407 degrees

m=2447 kg

as it is moving with constant speed, a=0

Friction force magnitude=m*(g*sin(theta)-a)=2447*(9.8*sin(7.407)-0)=3091.5 N

it will be directed upwards along the slope.

part b:

m=1576 kg

x=0.11

theta=arctan(0.11)=6.2772 degrees

a=-3.6 m/s^2(as it is slowing down)

Friction force magnitude=m*(g*sin(theta)-a)=1576*(9.8*sin(6.2772)-(-3.6))=7362.31 N

it will be directed upwards along the slope.

part c:

m=1379 kg

x=0.32

theta=17.7446 degrees

a=-1 m/s^2(as it is speeding up)

Friction force magnitude=m*(g*sin(theta)-a)=1379*(9.8*sin(17.7446)-(-1))=5497.78 N

it will be directed upwards along the slope.

part d:

m=1834 kg

x=0.11

theta=6.2772 degrees

a=-3.5 m/s^2 (as it is speeding up)

Friction force magnitude=m*(g*sin(theta)-a)=1834*(9.8*sin(6.2772)-(-3.5))=8384.16 N

it will be directed upwards along the slope.

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