A traveler pulls on a suitcase strap at an angle 36° above the horizontal. If 908 J of work are done by the strap while moving the suitcase a horizontal distance of 15 m, what is the tension in the strap?

T = 74.8 N

Explanation:

Given:

- The angle with the horizontal θ = 36°

- The tension in the strap T

- The amount of work done W = 908 J

- The distance travelled by the suitcase s = 15 m

Find:

what is the tension in the strap?

Solution:

- From definition we have that the work-done by a force F is given by the dot product of the force vector and the displacement vector as expressed below:

                                        W = F.s

-  The scalar form of the work done includes the component of force F along the direction of displacement as follows:

                                       W = F*s*cos(θ)

- Now we can calculate the Tension force T from the given data as follows:

                                       W = T*s*cos(θ)

                                       T = W / (s*cos(θ))

                                       T = 908 / (15*cos(36))

                                      T = 74.8 N

f=gm1m2/d^2 it d

f is force g is gravitaiton constant, m1 is mass of first object, m2 is mass of second object. d is distance between two object

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.