Question

10 formula 1 point a box has a mass of 3 kilograms. the box has an initial velocity of 8 meters per second. the box moves across the floor for 5 meters before coming to a stop due to the friction force acting on the box. determine the coefficient of friction for the box. hint: you can either use the work kinetic energy theorem or newton’s second law and kinematics. answer previous next

Explanation:

Step1: Recall Work - Kinetic Energy Theorem

The work - kinetic energy theorem states that the net work done on an object is equal to the change in its kinetic energy. The formula is \(W_{net}=\Delta KE = KE_f - KE_i\). Since the box comes to a stop, \(v_f = 0\), so \(KE_f=\frac{1}{2}mv_f^2 = 0\) and \(KE_i=\frac{1}{2}mv_i^2\). The net work done on the box is done by the frictional force. The work done by a force \(F\) over a displacement \(d\) is \(W = Fd\cos\theta\). For the frictional force, \(F_f=\mu_k N\) (kinetic friction, since the box is moving), and the normal force \(N\) on a horizontal surface is equal to the weight of the box, \(N = mg\). The angle between the frictional force and the displacement is \(180^{\circ}\) (because friction opposes motion), so \(\cos\theta=- 1\).

So, \(W_{net}=F_f d\cos\theta=-\mu_k mgd\) (the negative sign indicates that the frictional force does negative work as it opposes the motion). The change in kinetic energy is \(\Delta KE=0-\frac{1}{2}mv_i^2=-\frac{1}{2}mv_i^2\).

Step2: Set Work Equal to Change in KE

Set \(W_{net}=\Delta KE\):

$$-\mu_k mgd=-\frac{1}{2}mv_i^2$$

We can cancel out the negative signs and the mass \(m\) from both sides of the equation (since \(m
eq0\)):

$$\mu_k gd=\frac{1}{2}v_i^2$$

Now, solve for \(\mu_k\):

$$\mu_k=\frac{v_i^2}{2gd}$$

Step3: Substitute Values

We know that \(v_i = 8\space m/s\), \(g = 9.8\space m/s^2\), and \(d = 5\space m\). Substitute these values into the formula:

$$\mu_k=\frac{(8)^2}{2\times9.8\times5}=\frac{64}{98}\approx0.653$$

Answer:

The coefficient of friction \(\mu_k\approx0.65\) (or more precisely \(\frac{32}{49}\approx0.653\))