Questions: A contestant in a winter games event pushes a 40.0 kg block of ice across a frozen lake as shown in the figure below. (i) The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03. (For each answer, enter a number.) (a) Calculate the minimum force F (in N) he must exert to get the block moving. N (b) What is its acceleration (in m/s^2) once it starts to move, if that force is maintained? m/s^2

Solution Steps

Before the block starts moving, the forces acting on it are:

• The force of gravity downwards: \(mg\)
• The normal force upwards: \(N\)
• The applied force \(F\) at an angle of 25° above the horizontal.
• The static friction force opposing the horizontal component of \(F\): \(f_s\)

The block will start moving when the horizontal component of the applied force \(F\) overcomes the maximum static friction force.

The maximum static friction force is given by: \(f_{s,max} = \mu_s N\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force.

The normal force \(N\) is equal to the difference between the weight of the block and the vertical component of the applied force: \(N = mg - F\sin(25°)\)

Therefore, \(f_{s,max} = \mu_s (mg - F\sin(25°))\)

The horizontal component of the applied force \(F\) must be equal to the maximum static friction force to initiate motion. \(F\cos(25°) = f_{s,max}\) \(F\cos(25°) = \mu_s (mg - F\sin(25°))\)

Substituting the given values, we get: \(F\cos(25°) = 0.1(40.0 \times 9.8 - F\sin(25°))\) \(F(0.9063) = 39.2 - 0.0423F\) \(0.9486F = 39.2\) \(F = \frac{39.2}{0.9486}\) \(F \approx 41.32\) N

Once the block starts moving, the friction force changes from static friction to kinetic friction. The kinetic friction force is given by: \(f_k = \mu_k N = \mu_k (mg - F\sin(25°))\)

The net force acting on the block in the horizontal direction is: \(F_{net} = F\cos(25°) - f_k = F\cos(25°) - \mu_k(mg - F\sin(25°))\) Using Newton's second law, \(F_{net} = ma\), we can find the acceleration: \(a = \frac{F_{net}}{m} = \frac{F\cos(25°) - \mu_k(mg - F\sin(25°))}{m}\)

Substituting the values: \(a = \frac{41.32\cos(25°) - 0.03(40.0 \times 9.8 - 41.32\sin(25°))}{40.0}\) \(a = \frac{37.43 - 0.03(392 - 17.4)}{40.0}\) \(a = \frac{37.43 - 11.238}{40.0}\) \(a = \frac{26.192}{40.0}\) \(a \approx 0.655\) m/s²

Final Answer