Class Management I Help Hw #7 (Sece 5.1-2, Sec 7.1-75) Begin Date: 4/12/2015 12:

Question

Class Management I Help Hw #7 (Sece 5.1-2, Sec 7.1-75) Begin Date: 4/12/2015 12:00:00 AM Due Date: 4/27/2017 11:59:00 PM End Date: 5/4/2017 12:00:00 (10%) Problem 8: A shopper pushes a grocery cart for a distance 21.5 m at constant speed on level ground, against a 47.5 N frictional force. He pushes in a direction 24.5 below the horizontal Go A 20% Part (a) What is the work done on the cart by friction, in joules? Grade Sum Deductions Potential sino cos0 tan0 7 8 9 Submission Attempts re S 4 5 6 cotano asin0 acos0 4% per att detailed vie atan0 acotan0 sinh0 1 2 3 cosh0 tanh0 cotanh0 END vo Degrees Radians BACKSPACE Submit Hint Feedback I give up! Hints: 2% deduction per hint. Hints remaining: 1 Feedback: 2% deduction per feedback. A 20% Part (b) What is the work done on the cart by the gravitational force, in joules? A 20% Part (c What is the work done on the cart by the shopper, in joules? A 20% Part (d) Find the magnitude of the force, in newtons, that the shopper exerts on the cart. 20% Part (e) What is the total work done on the cart, in joules? All content 2017 Expert TA, LLC

Explanation / Answer

d = 21.5 m ; Ff = 47.5 N ; theta = 24.5 deg

a)The work done by the frictional force is

Wf = Ff d cos180

Wf = 47.5 x 21.5 x cos180 = - 1021.25 J

Wf = -1021.25 J

b)The work done by the gravity is

Wg = F d cos90 = 0

Wg = 0 J

c)since the cart is moving with constant speed,

Ws = + 1021.25 J

d)W = F d cos(theta)

F = W/d cos(theta)

F = 1021.25/21.5 x cos24.5 = 52.2 N

Hence, F = 52.2 N

e)Net work done is

Wnet = -1021.25 + 1021.25 = 0

Hence, Wnet = 0 J

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