Question

A block of mass \(20.0 \mathrm{~kg}\) supported by a vertical massless cable is initially at rest. The block is then pulled upward with a constant acceleration of \(2.32 \mathrm{~m} / \mathrm{s}^{2}\) a) What is the tension in the cable? b) What is the net force acting on the mass? c) What is the speed of the block after it has traveled \(2.00 \mathrm{~m}\) ?

Short answer

Answer: The tension in the cable is \(242.6 \mathrm{~N}\), the net force acting on the block is \(46.4 \mathrm{~N}\), and the speed of the block after traveling 2.00 m is \(3.05 \mathrm{~m/s}\).

Step-by-step solution

Calculate the gravitational force acting on the block

We first have to determine the gravitational force acting on the block, which is given by \(F_{g} = m \times g\), where \(m\) is the mass of the block and \(g = 9.81 \mathrm{~m} / \mathrm{s}^{2}\) is the gravitational acceleration. \(F_{g} = (20.0 \mathrm{~kg}) \times (9.81 \mathrm{~m/s^2}) = 196.2 \mathrm{~N}\)

Determine the tension in the cable using Newton's second law

According to Newton's second law of motion, the net force acting on an object is the product of its mass and acceleration, \(F_{net}=m\times a\). The net force acting on the block is the difference between the tension in the cable, \(F_{T}\), and the gravitational force, \(F_{g}\): \(F_{net} = F_{T} - F_{g}\). Plugging in the given acceleration and mass: \(F_{net} = (20.0 \mathrm{~kg}) \times (2.32 \mathrm{~m/s^2}) = 46.4 \mathrm{~N}\) Now, we can solve for the tension in the cable: \(F_{T} = F_{net} + F_{g} = 46.4 \mathrm{~N} + 196.2 \mathrm{~N} = 242.6 \mathrm{~N}\) So, the tension in the cable is \(242.6 \mathrm{~N}\).

Confirm the net force acting on the block

We have already calculated the net force in step 2, which was \(46.4 \mathrm{~N}\). Therefore, the net force acting on the block is \(46.4 \mathrm{~N}\).

Calculate the speed of the block after traveling 2.00 m

We will use the kinematic equation \(v^2 = u^2 + 2as\) to find the speed of the block after traveling 2.00 m. In this equation, \(v\) is the final speed, \(u\) is the initial speed (which is 0 because the block is initially at rest), \(a\) is the acceleration, and \(s\) is the distance traveled. \(v^2 = (0) + 2 \times (2.32 \mathrm{~m/s^2}) \times (2.00 \mathrm{~m})\) \(v^2 = 9.28 \mathrm{~m^2/s^2}\) Now, finding the square root to obtain the speed: \(v = \sqrt{9.28 \mathrm{~m^2/s^2}} = 3.05 \mathrm{~m/s}\) So, the speed of the block after traveling 2.00 m is \(3.05 \mathrm{~m/s}\).