A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer? A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer?

Step by step solution

01

Calculate the gravitational potential energy at the top

To calculate the potential energy, use the formula \(PE = mgh\), where \(m = 4.0 \, kg\), and \(h\) is the vertical distance from the top of the incline to the bottom. Since we have an inclination angle of \(15^{\circ}\) and the length of the incline is \(8.0 \, m\), we can find \(h\) as: \(h = L \sin(\theta) = 8.0\sin (15^{\circ})\)

02

Calculate the work done by gravity Wg

To calculate the work done by gravity on the block, we use the gravitational force acting on the block \(mg\) and the component of the force parallel to the incline \(mg\sin(\theta)\). The work done by gravity can be given as \(W_g = mg\sin(\theta) \times L\). Then we plug in the values: \(m = 4.0 \, kg\), \(g = 9.81 \, m/s^{2}\), \(\theta = 15^{\circ}\), and \(L = 8.0 \, m\)

03

Calculate the work done by tension Wt

The block stops at the bottom of the incline, so the final kinetic energy is \(0 J\). Using the work-energy theorem, we find the work done by tension across the remaining distance of the incline after cord is grasped (that is, \(3.0 \, m\)): $$W_{net} = W_g + W_t = \Delta KE = KE_f - KE_i = 0 - 0 = 0$$ Since we calculated \(W_g\) in step 2, we can find the work done by tension \(W_t\) from the above equation.

04

Calculate the tension in the cord T

To find the constant tension, we can divide the work done by tension \(W_t\) by the remaining distance of the incline, that is \(L_{remaining} = 3.0 \, m\), $$T = \frac{W_t}{L_{remaining}}$$ 2nd Method: Newton's Laws and constant acceleration

05

Set up a free-body diagram and write down Newton's second law components

We consider the block moving along the incline, and there are two forces acting on the block: 1. Gravitational force \(mg\) acting downward 2. Tension \(T\) acting along the incline (opposing the motion) After grasping the cord, the net force acting on the block along the incline is \(F_{net} = T - mg\sin(\theta)\). Using Newton's second law, \(F = ma\), we have: $$T - mg\sin(\theta) = ma$$

06

Calculate the stopping distance

Since the block comes to rest at the bottom of the incline, we can calculate the distance it takes to come to a stop after the cord is grasped using a constant acceleration equation, where \(s = L_{remaining} = 3.0 \, m\), \(v_f = 0 \, m/s\) and \(v_i\) is the initial velocity when the cord is grasped (which we can calculate using work-energy theorem as \(\frac{1}{2}mv_i^2 = mgh - mg\sin(\theta) \times 5.0\)) We can use the equation: \(v_f^2 = v_i^2 + 2as\) Solve for acceleration \(a\).

07

Calculate the tension T

Now we have the acceleration \(a\), we can plug it back into the equation derived from Newton's second law in Step 1: $$T = ma + mg\sin(\theta)$$ The constant tension T can be calculated using the values obtained from the above steps. Compare the values from both methods and choose the preferred method.

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