Can you please explain the answers. Thanks a lot Long Problems: (30 points) Show

Question

Can you please explain the answers. Thanks a lot Long Problems: (30 points) Show all work, including showing any formulas used before you put any numbers in. Show all units. 1. (151 A truck with failed brakes accelerates (in the +x direction) down a slope angled at phi = 8? above the horizontal. The truck has a mass of 40,000 kg, and the force labeled F air is 20,000 N. Note that the truck is traveling in the +x direction in the picture. a. (3) Write down Newton's 2nd Law in the x direction. (3) Write down Newton's 2nd Law in the y direction. c. (2) Find the normal force N. d. (3) Find the acceleration of the truck. e. (2) How much work is done by the normal force if the truck has a displacement of 100m down the slope? f. (2) How much work is done by the weight if the truck has a displacement of 100m down the slope?

Explanation / Answer

1)

Q = 8 degrees

a)

Newton's law in x-direction:

Fnet,x = mg*sinQ - Fair = m*ax

where ax = acceleration in x direction

m = 40000 kg

b)

Newton's 2nd law in y-direction

Fnet,y = mg*cosQ - N = m*ay

c)

Now from part(b),

Fnet,y = mg*cosQ - N = m*ay

But the body doesnt move in y - direction , so ay = 0

So, mg*cosQ - N = 0

So, N = mg*cosQ

= 40000*9.8*cos(8 degrees)

= 3.88*10^5 N <---------answer

d)

acceleration of truck = ax

From part (a),

Fnet,x = mg*sinQ - Fair = m*ax

So, ax =( mg*sinQ-Fair)/m

= (40000*9.8*sin(8 degrees)-20000)/40000

= 0.864 m/s2 <--------answer

e)

Work done by normal force = 0 <------ as Normal force is perpendicular to the direction of motion

f)

Work done by weight = mg*sinQ*(displacement) = mg*sinQ*100

So, W = 40000*9.8*sin(8 degrees)*100 = 5.46*10^6 J <------------answer

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