Question

two students push a 290 kg piano across the floor. kathy pushes with 280 n forward while matt pushes with 240 n forward. the piano accelerates at 0.20 m/s² forward. (a) what is the coefficient of kinetic friction? (b) how long will it take the piano to stop moving after pushing it for 6.2 s from rest?

Explanation:

Part (a)

Step 1: Find the net force

The total force applied by Kathy and Matt is \( F_{\text{applied}} = 280\,\text{N} + 240\,\text{N} = 520\,\text{N} \). Using Newton's second law \( F_{\text{net}} = ma \), where \( m = 290\,\text{kg} \) and \( a = 0.20\,\text{m/s}^2 \), we get \( F_{\text{net}} = 290\,\text{kg} \times 0.20\,\text{m/s}^2 = 58\,\text{N} \).

Step 2: Find the frictional force

The frictional force \( F_f \) opposes the applied force, so \( F_f = F_{\text{applied}} - F_{\text{net}} = 520\,\text{N} - 58\,\text{N} = 462\,\text{N} \).

Step 3: Find the normal force

The normal force \( F_N \) on the piano (on a horizontal surface) is equal to its weight: \( F_N = mg \), where \( g = 9.8\,\text{m/s}^2 \). So \( F_N = 290\,\text{kg} \times 9.8\,\text{m/s}^2 = 2842\,\text{N} \).

Step 4: Calculate the coefficient of kinetic friction

The formula for kinetic friction is \( F_f = \mu_k F_N \), so \( \mu_k = \frac{F_f}{F_N} = \frac{462\,\text{N}}{2842\,\text{N}} \approx 0.16 \).

Step 1: Find the velocity after pushing

Using \( v = u + at \), where \( u = 0\,\text{m/s} \), \( a = 0.20\,\text{m/s}^2 \), and \( t = 6.2\,\text{s} \), we get \( v = 0 + 0.20\,\text{m/s}^2 \times 6.2\,\text{s} = 1.24\,\text{m/s} \).

Step 2: Find the deceleration due to friction

When the students stop pushing, the net force is the frictional force. Using \( F_{\text{net}} = F_f = ma' \), we get \( a' = \frac{F_f}{m} = \frac{462\,\text{N}}{290\,\text{kg}} \approx -1.59\,\text{m/s}^2 \) (negative because it's deceleration).

Step 3: Find the time to stop

Using \( v = u + a't' \), where \( v = 0\,\text{m/s} \), \( u = 1.24\,\text{m/s} \), and \( a' = -1.59\,\text{m/s}^2 \), we solve for \( t' \): \( 0 = 1.24\,\text{m/s} + (-1.59\,\text{m/s}^2)t' \). So \( t' = \frac{1.24\,\text{m/s}}{1.59\,\text{m/s}^2} \approx 0.78\,\text{s} \).

Answer:

The coefficient of kinetic friction is approximately \( \boldsymbol{0.16} \).

Part (b)