.ppc- Tut Exercise A 197-g block is pressed against a spring of force constant 1.42 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0 to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions. (a) if the ramp exerts no friction force on the block (b) if the coefcient of kinetic friction is 0.380 Part 1 of 5 - Conceptualize Part of this problem describes a situation that can be duplicated in a laboratory experiment. We expect a distance of a few meters if the ramp exerts no friction and perhaps half the distance if we consider force due to friction. Part 2 of 5 - categorize We must compute the spring energy before and alter launch. We will use the nonisolated system energy model to be sure that we keep track of the energy from all of the sources. Part 3 of 5 -Analyze We consider gravitational energy, elastic energy, kinetic energy, and internal energy. The spring is initially compressed by xi = 0.100 m. The block travels up the ramp 3 distance d. The work that the spring does on the block is given by the following equation. W; = in? kxf2 = %kxi2 o = lloqz. 2 Gravity does work on the block specified in the equation below. There is no friction. wg = mgd cos(90 + 60 y f} 60 ) mgd sin( 60 y /) 60 ) (b) Within the system, friction transforms kinetic energy into internal energy. For the internal energy, we have AEint = fkd = (ukn)d = MK mg cos 60 60 .) d. For the total energy, we have the following equation. Ws + Wg - AEint = 0 For the change in internal energy of the system due to friction, we have the following. AEint = MK mg cos 60 | 60 .) d Substituting into the total energy equation above, gives us 1kx;2 - mgd sin 60 60 0 - MK COS 60 60 . d = 0. Part 5 of 5 - Analyze Now we substitute the expression we found above for AEint and the values we know into the total energy equation. This gives us the following. 2 1.42 x 103 N/m 10e- 2 m) - .197 kg)(9.80 m/s2) a (sin 60 10 ) 1.97 kg)(9.80 m/52 ) cos 60 1. ) d = 0 Solving the equation above for d, the distance the block moves up the ramp in the presence of friction, gives d = m. Submit Skip (you cannot come back). Viewing Saved Work Revert to Last Response Submit Answer View Previous Question Question 16 of 19 View Next