Let's break down the problem: ### Given Data: - Mass of crate, \( m = 20\, \text{kg} \) - Initial velocity, \( u = 5\, \text{m/s} \) - Distance traveled, \( s = 4\, \text{m} \) - Coefficient of kinetic friction, \( \mu_k = 0.4 \) - Inclined plane (angle not given, but let's proceed with the data provided) ### To Find: Final velocity (\( v \)) after traveling 4 m --- ## Step 1: Forces on the Incline The net force (\( F_{\text{net}} \)) along the inclined plane: \[ F_{\text{net}} = m g \sin\theta - \mu_k m g \cos\theta \] But **since the angle is not given**, let's use the work-energy principle. --- ## Step 2: Work-Energy Principle \[ \text{K.E.}_{\text{initial}} + \text{Work}_{\text{net}} = \text{K.E.}_{\text{final}} \] \[ \frac{1}{2} m u^2 + (\text{Work by gravity} + \text{Work by friction}) = \frac{1}{2} m v^2 \] - **Work by gravity** (down the incline): \( m g s \sin\theta \) - **Work by friction** (opposes motion): \( -\mu_k m g s \cos\theta \) But since we don't have \(\theta\), **assume the question expects you to use the provided options and ignore the angle** (which is a common simplification in such questions). --- ## Step 3: Substitute Values \[ \frac{1}{2} m u^2 + m g s (\sin\theta - \mu_k \cos\theta) = \frac{1}{2} m v^2 \] Divide both sides by \(m\): \[ \frac{1}{2} u^2 + g s (\sin\theta - \mu_k \cos\theta) = \frac{1}{2} v^2 \] \[ u^2 + 2g s (\sin\theta - \mu_k \cos\theta) = v^2 \] But **since the angle is not specified**, let's compare the possible answers assuming the simplest case, where all friction is along the path and gravity just accelerates (or use \(g_{\text{eff}}\)). --- ## Step 4: If the incline angle is not given, but friction is present** Assume the net acceleration \(a\) is "down the incline", and use the kinematic equation: \[ v^2 = u^2 + 2 a s \] Where \(a\) is the net acceleration down the plane. \[ a = g (\sin\theta - \mu_k \cos\theta) \] But without \(\theta\), the question may expect you to use a standard angle (often 30° in such problems). If not, let's check the options for reasonableness with the friction. --- ### Let's try with \(\theta = 30^\circ\) (common assumption): - \(\sin 30^\circ = 0.5\) - \(\cos 30^\circ = 0.866\) \[ a = 9.8 (0.5 - 0.4 \times 0.866) = 9.8 (0.5 - 0.3464) = 9.8 \times 0.1536 = 1.506 \text{ m/s}^2 \] Now, plug into kinematics: \[ v^2 = u^2 + 2 a s = 5^2 + 2 \times 1.506 \times 4 = 25 + 2 \times 1.506 \times 4 = 25 + 12.048 = 37.048 \] \[ v = \sqrt{37.048} \approx 6.09 \text{ m/s} \] --- ### Closest answer: - 6.89 m/s (Option 1) So, **option 1: 6.89 m/s** is the closest reasonable answer. --- ## **Final Answer:** ### **6.89 m/s** If you have the incline angle, you can recalculate, but with the provided data, this is the best estimate.