Question

1. a box that weighs 2,000 n is accelerated uniformly over a horizontal surface at a rate of 8 m/s². the opposing force of friction between the box and the surface is 27.4 n.

(a) how much force is being exerted horizontally on the box?
(b) how much force would be required to lift the box upward with an acceleration of 8 m/s²? draw a fbd first!

1. a force of 20n accelerated a 9.0 kg wagon at 2m/s2 along a sidewalk.

(a) how large is the frictional force?
(b) what is the coefficient of friction?

Explanation:

Step1: Find mass of the box in problem 5

Use formula \(W = mg\), so \(m=\frac{W}{g}\). Given \(W = 2000N\) and \(g = 9.8m/s^{2}\), then \(m=\frac{2000}{9.8}\approx204.08kg\).

Step2: Solve part (a) of problem 5

By Newton's second law \(F_{net}=ma\). Let the applied force be \(F\). Then \(F - f=ma\). Substitute \(m = 204.08kg\), \(a = 8m/s^{2}\), \(f = 27.4N\). So \(F=ma + f=204.08\times8+27.4 = 1632.64+27.4=1660.04N\).

Step3: Solve part (b) of problem 5

For vertical motion, \(F - W=ma\). Substitute \(W = 2000N\), \(m = 204.08kg\), \(a = 8m/s^{2}\). Then \(F=W+ma=2000 + 204.08\times8=2000+1632.64 = 3632.64N\).

Step4: Solve part (a) of problem 6

By \(F_{net}=ma\). Let the frictional force be \(f\). Then \(F - f=ma\). Given \(F = 20N\), \(m = 9.0kg\), \(a = 2m/s^{2}\). So \(f=F - ma=20-9\times2=20 - 18=2N\).

Step5: Solve part (b) of problem 6

Since \(f=\mu N\) and on horizontal surface \(N = mg\) (\(m = 9.0kg\), \(g = 9.8m/s^{2}\), \(N=9\times9.8 = 88.2N\), \(f = 2N\)). Then \(\mu=\frac{f}{N}=\frac{2}{88.2}\approx0.023\).

Answer:

5(a). \(1660.04N\)
5(b). \(3632.64N\)
6(a). \(2N\)
6(b). \(0.023\)