Respuesta :
Answer:
Part a) Work done by friction force
W = -700 J
Part b) Work done by gravity
W = 0
Part c) Work done by shopper
W = 700 J
Part d) Force exerted by shopper
F = 38.6 N
Part e) Total work done
W = 0
Explanation:
Part a)
Work done by friction force is given as
[tex]W = -F_f (d)[/tex]
so we have
[tex]W = -(35)(20)[/tex]
[tex]W = -700 J[/tex]
Part b)
Since gravitational force is perpendicular to the displacement
so we have
[tex]W = Fd cos90[/tex]
[tex]W = 0[/tex]
Part c)
Work done by the shopper is same as that of work done by friction force
Because here trolley is moving with constant velocity
So net force on it is zero
[tex]W_{shopper} + W_{friction} = 0[/tex]
[tex]W_{shopper} = - W_{friction}[/tex]
[tex]W_{shopper} = 700 J[/tex]
Part d)
[tex]F_f + Fcos25 = 0[/tex]
[tex]-35 + Fcos25 = 0[/tex]
[tex]F = 38.6 N[/tex]
Part e)
Total work done is given as
[tex]W = W_{shopper} + W_{friction}[/tex]
[tex]W = 0[/tex]
The work done by zero net force acting on a body is zero
The required values are;
(a) The work done by friction = 700 J
(b) Work done by gravity, [tex]W_g[/tex] = 0
(c) The work done by the shopper = 35.0 N × 20.0 m = 700 J
(d) The force the shopper exerts is approximately 38.618 N
(e) The total work done on the cart is 0
The reason the above values are correct is as follows:
Known parameters:
The distance the shopper pushes the grocery cart, d = 20.0 m
Characteristics of the motion of the cart = Constant speed (zero acceleration)
Frictional force, [tex]F_f[/tex] = 35.0 N
The direction the cart is pushed, θ = 25.0° below the horizontal
The nature of the ground = Level ground
Let F represent the force with which the shopper pushes, we have;
(a) Work done = Force × Distance in the direction of force
The work done by friction, [tex]W_f[/tex] = [tex]F_f[/tex] × d
∴ [tex]W_f[/tex] = 35.0 N × 20.0 N = 700 J
The work done by friction = 700 J
(b) The distance moved in gravity direction (downwards) is zero, therefore;
Work done by gravity, [tex]W_g[/tex] = W × 0 = 0
Where;
W = The weight of the cart
Work done by gravity, [tex]W_g[/tex] = 0
(c) The work done by the shopper = Force applied by the shopper in the direction of motion of the cart × The distance the cart moves
Given that the shopper pushes the cart at constant speed, the acceleration of the cart in the direction of motion is zero
Therefore, the net force on the cart is zero
The force applied by the shopper in the direction of motion of the cart is equal to the frictional force = 35.0 N
∴ The work done by the shopper = Force applied by the shopper in the direction of motion of the cart × The distance the cart moves
The work done by the shopper = 35.0 N × 20.0 m = 700 J
(d) The force the shopper exerts, F, is given as follows;
F × cos (θ) - [tex]F_f[/tex] = m × a
The acceleration of the cart, a ≈ 0
∴ F × cos (θ) - [tex]F_f[/tex] = m × 0 = 0
F × cos (θ) = [tex]F_f[/tex]
∴ F × cos (25.0°) = 35.0 N
F = 35.0 N/(cos(25.0°) ≈ 38.618 N
The force the shopper exerts, F ≈ 38.618 N
(e) The total work done on the cart = The work done by the shopper + [tex]W_f[/tex]
∴ The total work done on the cart = 700 J + (-700 J) = 0
The total work done on the cart = 0
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