Physics HW Help?

Question

A box m = 53 kg is being pulled by a constant force F = 166 N at an angle of θ = 18 degrees. The initial speed of the box is zero.

A.) Write an expression for the work done by force F as the block moves a horizontal distance d.

B.) How much work, in joules, was done in moving the block 5.2 m?

C.) What is the speed of the box at d = 5.2 m if the surface is frictionless?

Ash · Accepted Answer

Considering the angle θ = 18° is above the horizontal

A) Force along the surface, Fx = Fcosθ

Work done, W = Fx*d
W = F cosθ * d

B) 
W = (166 N)cos18° * (5.2 m) = 821 J

C) ΔKE = W
KEf - KEi = W
½mv² - 0 = W
v² = 2W/m
v² = 2(821 J)/(53 kg)
v = 5.57 m/s

billrussell42 · Answer

what is the angle with respect to? assuming horizontal.

F = 160 cos 18 = 152.2

w = Fd= 152.2•d

w = 152.2•5.2 = 791 J

a = F/m = 152.2/53 = 2.87 m/s²
v² = v₀² + 2ad
v² = 0² + 2(2.87)(5.2)
v = 5.46 m/s

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