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
Favorite 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
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