Car on ramp 4 A 2,073-kg car is moving up a road with a slope (grade) of 32% at

Question

Car on ramp 4

A 2,073-kg car is 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., up the slope)? A 1,495-kg car is moving up a road with a slope (grade) of 35% while slowing down at a rate of 1.1 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., up the slope)? A 1,953-kg car is moving up a road with a slope (grade) of 13% while slowing down at a rate of 4.4 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., up the slope)? A 1,712-kg car is moving up a road with a slope (grade) of 24% while speeding up at a rate of 1.4 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., up the slope)?

Explanation / Answer

A) constant speed means accelaration a= 0 m/s^2

then net force Fnet = m*a = 0

m*g*sin(11) = fk = frictional force

fk = 2447*9.81*sin(11) = 4580.4 N

direction is in upward direction

B) accelaration a = 3.6 m/s^2

fk-m*g*sin(11)= m*a

fk = ma + mg*sin(11)

fk = 1576(3.6 + (9.81*sin(11)))= 8623.6 N upward direction

C) a = 1 m/s^2

m*g*sin(32) - fk = m*a

fk = 1379*(9.81*sin(32) - 1) = 5789.74 N upward direction

D) a = 3.5 m/s^2

m*g*sin(32) - fk = m*a

fk = 1834*(9.81*sin(11) - 3.5) = -2986.05 down the slope

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