A horizontal force of 200 N is applied to a \( 55-\mathrm{kg} \) cart across a \( 10-\mathrm{m} \) level surface. If the cart accelerates at \( 2.0 \mathrm{~m} / \mathrm{s}^{2} \), then what is the work done by the force of friction as it acts to retard the motion of the cart? 1) -1100 J 2) -900 J 3) -800 J 4) -700 J 5) -600 J

To find the work done by the force of friction, we first need to calculate the net force acting on the cart. The net force can be determined using Newton's second law: \[ F_{\text{net}} = m \times a = 55 \, \text{kg} \times 2.0 \, \text{m/s}^2 = 110 \, \text{N}. \] Now, we know the applied force is 200 N. The frictional force (F_friction) can be found using the relationship: \[ F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}. \] Plugging in the known values: \[ 110 \, \text{N} = 200 \, \text{N} - F_{\text{friction}} \] Solving for \( F_{\text{friction}} \): \[ F_{\text{friction}} = 200 \, \text{N} - 110 \, \text{N} = 90 \, \text{N}. \] Now, to find the work done by friction (W_friction): \[ W_{\text{friction}} = F_{\text{friction}} \times d \times \cos(\theta), \] where \( d = 10 \, \text{m} \) and \( \theta = 180^\circ \) (since friction acts in the opposite direction of motion). Thus, \( \cos(180^\circ) = -1 \). So, \[ W_{\text{friction}} = 90 \, \text{N} \times 10 \, \text{m} \times (-1) = -900 \, \text{J}. \] Therefore, the answer is **-900 J (Option 2)**. Now you know how the math pieces fit together to find the work done by friction!