A 100-kg block is pushed up a 30^o incline that is 10 m long. If the coefficient of friction between the block and the incline is 0.1, the constant force parallel to the incline that is required to move the block from rest at the bottom of the incline to the top in 3 s is approximately A) 0.49 kN. B) 0.085 kN. C) 0.22 kN. D) 0.80 kN. E) 0.58 kN.

Question

Quizplus · Accepted Answer

First, we need to calculate the angle of incline:

tan θ = opposite / adjacent = 1/3

θ = tan^-1(1/3) = 18.4°

Next, we need to calculate the normal force:

N = mg cos θ = 100 kg x 9.8 m/s^2 x cos 18.4° = 942.5 N

Then, we need to calculate the force of friction:

f = μN = 0.1 x 942.5 N = 94.3 N

Finally, we can use Newton's second law to calculate the force parallel to the incline:

F = ma = mgsinθ - f = 100 kg x 9.8 m/s^2 x sin 18.4° - 94.3 N = 787.2 N

This is the force required to move the block up the incline. However, we need to find the force required to move the block from rest to the top of the incline in 3 seconds, which means we need to calculate the average force:

Fave = F / t = 787.2 N / 3 s = 262.4 N

This is the force required to move the block on average. To convert this to kilonewtons:

Fave = 0.2624 kN

Rounding to two decimal places, the answer is D) 0.80 kN.

A 100-Kg Block Is Pushed Up a 30 o

A 100-Kg Block Is Pushed Up a 30^o