Respuesta :
Answer:
Tension, T = 74.82 N
Explanation:
It is given that,
Angle between the suitcase and horizontal, [tex]\theta=36^{\circ}[/tex]
Work done, W = 908 J (let)
Distance covered, d = 15 m (let)
We know that the work done by an object is given by the formula as :
[tex]W=Fd\ cos\theta[/tex]
here, F = T
[tex]W=Td\ cos\theta[/tex]
[tex]T=\dfrac{W}{d\ cos\theta}[/tex]
[tex]T=\dfrac{908}{15\times cos(36)}[/tex]
T = 74.82 N
So, the tension in the strap is 74.82 N. Hence, this is the required solution.
Answer:
74.79N
Explanation:
The question is incomplete. Here is the complete question.
" A traveler pulls on a suitcase straps at an angle 36° above the horizontal. If 908J of work are done by the strap while moving the suitcase a a horizontal distance of 15m, what is the tension in the strap?"
Work is said to be done if an applied force cause an object to move through a distance.
Work done = Force F × perpendicular distance d
Since the strap is pulled at ana angle above the horizontal,
Work done = F ×d cos(theta)
Given workdone = 908Joules
d = 15m
theta = 36°
908 = F × 15cos36°
F = 908/15cos36°
F = 908/12.14
F = 74.79N
The tension in the strap = 74.79N