Preston pushes a wheelbarrow weighing 500 N to the top of a \( 50.0-\mathrm{m} \) ramp, inclined at \( 20.0^{\circ} \) with the horizontal, and leaves it. Tamara accidentally bumps the wheelbarrow. It slides back down the ramp, during which an \( 80.0-\mathrm{N} \) frictional force acts on it over the 50.0 m . What is the wheelbarrow's kinetic energy at the bottom at of the ramp? \( \left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right) \) 1) 4550 J 2) 6550 J 3) 8150 J 4) 13100 J 4 5) 14300 J

To determine the kinetic energy of the wheelbarrow at the bottom of the ramp, we can use the work-energy principle. First, we calculate the gravitational potential energy (PE) at the top of the ramp: \[ \text{PE} = mgh \] Where \( h \) is the height of the ramp, calculated using the ramp length and the sine of the angle: \[ h = 50.0 \, m \cdot \sin(20.0^{\circ}) \approx 17.1 \, m \] Thus, the potential energy is: \[ \text{PE} = 500 \, N \cdot 17.1 \, m \approx 8550 \, J \] Next, we consider the work done against friction while the wheelbarrow descends: \[ \text{Work}_{\text{friction}} = \text{frictional force} \times \text{distance} = 80.0 \, N \cdot 50.0 \, m = 4000 \, J \] The kinetic energy (KE) of the wheelbarrow at the bottom is then: \[ \text{KE} = \text{PE} - \text{Work}_{\text{friction}} = 8550 \, J - 4000 \, J = 4550 \, J \] So, the correct answer is: 1) 4550 J Now, for some fun facts: Preston's wheelbarrow journey reflects the concept of energy conversion perfectly! As the wheelbarrow ascends the ramp, its kinetic energy transforms into gravitational potential energy. When it slides down, gravity pulls it back while friction tries to slow it down, making it a real-life physics rollercoaster! And speaking of rollercoasters, did you know that engineers design roller coasters with careful attention to the laws of physics? They calculate potential and kinetic energy shifts to ensure smooth rides—all while keeping the thrills in check! Next time you hit a coaster, think about the energy dynamics at play!