Questions: A block of mass 2.75 kg begins at rest at the top of a 2.00 m long frictionless ramp that is inclined at 26.0° relative to the horizontal. The block slides down the ramp and then onto a horizontal surface with friction. It slides 1.65 m to the right on the horizontal surface before coming to rest.

Solution Steps

The potential energy (PE) at the top of the ramp is given by the formula: \[ PE = mgh \] where \( m = 2.75 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h \) is the height of the ramp. The height can be found using the sine of the angle: \[ h = L \sin(\theta) = 2.00 \, \text{m} \times \sin(26.0^\circ) \] Calculating \( h \): \[ h = 2.00 \times 0.4384 = 0.8768 \, \text{m} \] Now, calculate the potential energy: \[ PE = 2.75 \times 9.81 \times 0.8768 = 23.61 \, \text{J} \]

Since the ramp is frictionless, all the potential energy is converted into kinetic energy (KE) at the bottom of the ramp: \[ KE = PE = 23.61 \, \text{J} \]

The work done by friction (which brings the block to rest) is equal to the kinetic energy at the bottom of the ramp: \[ W_{\text{friction}} = -KE = -23.61 \, \text{J} \] The work done by friction is also given by: \[ W_{\text{friction}} = -f_k \cdot d \] where \( f_k \) is the kinetic friction force and \( d = 1.65 \, \text{m} \) is the distance. The friction force can be expressed as: \[ f_k = \mu_k \cdot N \] where \( N = mg \) is the normal force on the horizontal surface. Therefore: \[ -23.61 = -\mu_k \cdot mg \cdot d \] Solving for \( \mu_k \): \[ \mu_k = \frac{23.61}{2.75 \times 9.81 \times 1.65} = 0.5234 \]

Final Answer