**Problem Statement:**Coefficient of kinetic friction between the mass and the incline is 0.2. Find the minimum work done to slide the mass from the bottom to the top of the incline.**Diagram Explanation:**- The diagram shows a 10 kg block on an inclined plane with an angle of \( \theta = 30^\circ \).- The height of the incline is given as \( h = 4 \, \text{m} \).- A force, \( f_1 = 50 \, \text{N} \), is applied parallel to the incline.- The coefficient of kinetic friction is 0.2.**Physics Concepts:**The work done to move an object up an incline is calculated considering gravitational forces, frictional forces, and any applied force. The work done \( W \) can be expressed as:\[ W = mgh + f_f d \]where:- \( m \) is mass of the object,- \( g \) is the acceleration due to gravity \( (9.8 \, \text{m/s}^2) \),- \( h \) is the height of the incline,- \( f_f \) is the frictional force,- \( d \) is the distance along the incline.The distance \( d \) can be found using trigonometry:- \( d = \frac{h}{\sin(\theta)} \).The frictional force \( f_f \) is calculated as:\[ f_f = \mu_k \cdot N \]Where \( N \) is the normal force.The normal force \( N \) on an incline is:\[ N = mg \cos(\theta) \]Therefore, the work done against friction is:\[ W_{\text{friction}} = f_f \cdot d = \mu_k \cdot mg \cos(\theta) \cdot d \]Finally, add these components to find the total work done to overcome gravity and friction.

Name: **Problem Statement:**Coefficient of kinetic friction between the mas
Uploaded: 2024-03-20T06:57:25.000Z
Channel: Bartleby
Description: **Problem Statement:**Coefficient of kinetic friction between the mass and the incline is 0.2. Find the minimum work done to slide the mass from the bottom to the top of the incline.**Diagram Explanation:**- The diagram shows a 10 kg block on an inc

Answered: Image /qna-images/answer/94009847-4eec-44de-921b-a7f765a19ccf.jpgGiven, m=10kg θ=30° μ=0.2 g=10m/s2 Height=h=4m The contact force is Fc is the resultant of frictional force and normal force. From the free body diagram, Normal force, N=mgcosθ Frictional force=μN=μmgcosθ N and Fr are perpendicular to each other. The contact force is, Fc=N2+Fr2 Since contact force is perpendicular to height h, work done=Fc.hNormal force, N=mgcosθ=50N (given in figure) Frictional force=μN=μmgcosθ=0.2x50=10N The contact force is, Fc=N2+Fr2=502+102=50.99N Minimum Work done=Fc x h=50.99 x 4=203.96 J