A box of mass \( m = 20.0 \, \text{kg} \) is pulled up a ramp that is inclined at an angle \( \theta = 15.0^\circ \) with respect to the horizontal. The coefficient of kinetic friction between the box and the ramp is \( \mu_k = 0.335 \), and the rope pulling the box is parallel to the ramp. If the box accelerates up the ramp at a rate of \( a = 2.89 \, \text{m/s}^2 \), calculate the tension \( F_T \) in the rope. Use \( g = 9.81 \, \text{m/s}^2 \) for the acceleration due to gravity.\( F_T = \) \_\_\_\_ N**Diagram Explanation:**The diagram shows a box on an inclined ramp. The box is being pulled by a rope parallel to the ramp. The angle \( \theta \) is shown between the horizontal and the surface of the ramp. The direction of the acceleration \( \vec{a} \) is indicated by an arrow pointing up the ramp. The coefficient of kinetic friction \( \mu_k \) is also noted on the diagram.

Name: A box of mass \( m = 20.0 \, \text{kg} \) is pulled up a ramp that is
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: A box of mass \( m = 20.0 \, \text{kg} \) is pulled up a ramp that is inclined at an angle \( \theta = 15.0^\circ \) with respect to the horizontal. The coefficient of kinetic friction between the box and the ramp is \( \mu_k = 0.335 \), and the rop

Given Mass of the box m=20.00 kg The angle of inclination θ=150 Coefficient of kinetic friction μk=0.335 The acceleration of the ramp a=2.89 m/s2 The acceleration due to gravity g=9.8m/s2 The free body diagram of the block is as shown below N is the normal force mg is the weight of the block fk is the frictional force F is the net force on the block F=ma=20.0 kg2.89 m/s2F=57.8 N From free body diagram N=mgcosθFT-mgsinθ+fk=Ffrictional force fk=μkN=μkmgcosθFT=F+mgsinθ+μkmgcosθ The tension in the rope is FT=57.8N+20.0 kg9.8 m/s2sin150+0.33520.0 kg9.8 m/s2cos150FT=171.95 NAnswer : 171.95 N