A 0.75-kg block slides on a rough horizontal table top.Just before it hits a horizontal ideal spring its speed is 3.5 m/s.It compresses the spring 5.7 cm before coming to rest.If the spring constant is 1200 N/m, the thermal energy of the block and the table top must have:
A)not changed
B)decreased by 1.9 J
C)decreased by 2.6 J
D)increased by 1.9 J
E)increased by 2.6 J

Question

Quizplus · Accepted Answer

The thermal energy of the block and the table top increases due to the work done against friction. The kinetic energy of the block is converted into elastic potential energy stored in the spring and thermal energy due to friction. The increase in thermal energy can be calculated by finding the initial kinetic energy of the block and the potential energy stored in the spring. The initial kinetic energy (KE) is given by $ \frac{1}{2}mv^2 $, where $ m = 0.75 $ kg and $ v = 3.5 $ m/s. The potential energy (PE) stored in the spring is given by $ \frac{1}{2}kx^2 $, where $ k = 1200 $ N/m and $ x = 0.057 $ m (since 5.7 cm = 0.057 m). The difference between the initial kinetic energy and the potential energy stored in the spring is the increase in thermal energy.Calculating the initial kinetic energy:$$ KE = \frac{1}{2} \times 0.75 \times (3.5)^2 = 4.59375 \, \text{J} $$Calculating the potential energy stored in the spring:$$ PE = \frac{1}{2} \times 1200 \times (0.057)^2 = 1.9442 \, \text{J} $$The increase in thermal energy is the difference between the initial kinetic energy and the potential energy stored in the spring:$$ \Delta E_{thermal} = KE - PE = 4.59375 - 1.9442 = 2.64955 \, \text{J} $$This is approximately 2.6 J, so the thermal energy of the block and the table top has increased by 2.6 J.

A 075-Kg Block Slides on a Rough Horizontal Table Top