Question

1. a 1200 kg car travels at a constant speed on a hill of radius 48 m as shown.

v = 14 m/s

r = 48 m

a. draw a free - body of the force acting on the car. (2 marks)

b. if the car travels at the top of the hill at 14 m/s, what is the normal force exerted by the road on the car? (3 marks)

c. what is the minimum speed that the car can have as it passes the top of the hill before losing contact with the road (becoming weightless)? (3 marks)

1. a 2.5 kg pendulum bob that is 1.6 m long is pulled horizontal and then released.

a. what is the speed of the bob at the bottom of its path? (2 marks)

b. the maximum tensional force of the string on the pendulum bob is 70 n. will the pendulum bob string break? if so where in its motion would it do so? (4 marks)

Explanation:

7.b

Step1: Analyze forces and apply Newton's second law

At the top of the hill, the net force \(F_{net}=mg - F_N\) provides the centripetal force \(F_c=\frac{mv^{2}}{R}\). So, \(mg - F_N=\frac{mv^{2}}{R}\).

Step2: Solve for the normal force \(F_N\)

Rearrange the formula \(F_N = mg-\frac{mv^{2}}{R}\). Substitute \(m = 1200\space kg\), \(g = 9.8\space m/s^{2}\), \(v = 14\space m/s\), and \(R=48\space m\).

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Step1: Determine the condition for losing contact

When the car loses contact with the road (\(F_N = 0\)), the centripetal force is provided solely by gravity. So, \(mg=\frac{mv_{min}^{2}}{R}\).

Step2: Solve for the minimum speed \(v_{min}\)

Cancel out \(m\) from both sides of the equation \(mg=\frac{mv_{min}^{2}}{R}\), getting \(v_{min}=\sqrt{gR}\). Substitute \(g = 9.8\space m/s^{2}\) and \(R = 48\space m\).

$$v_{min}=\sqrt{9.8\times48}\approx\sqrt{470.4}\approx21.7\space m/s$$

Step1: Apply the conservation of mechanical energy

The initial potential energy \(U = mgh\) (where \(h = L\), the length of the pendulum) is converted into kinetic energy \(K=\frac{1}{2}mv^{2}\) at the bottom. So, \(mgL=\frac{1}{2}mv^{2}\).

Step2: Solve for the speed \(v\)

Cancel out \(m\) from both sides of the equation \(mgL=\frac{1}{2}mv^{2}\), getting \(v=\sqrt{2gL}\). Substitute \(g = 9.8\space m/s^{2}\) and \(L = 1.6\space m\).

$$v=\sqrt{2\times9.8\times1.6}=\sqrt{31.36}= 5.6\space m/s$$

Answer:

The normal force exerted by the road on the car is \(6860\space N\)

7.c