Question

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1) A 500 N force directed 30 degrees above the horizontal is used to pull a 50 kg sled across the ground. The coefficient of friction between the sled and the ground is 0.3. The sled is pulled 5 meters. a) What is the work done by the pulling force?

Answer 1

Fexe Bedy dingrem nee N F-S0ON

Since work done is given by,

W = F*d*cos$\Theta$

where, F = applied force

d = displacement

$\Theta$ = angle between displacement and force

a.)

here, F = pilling force = 500 N

d = 5 m

$\Theta$ = 30 deg

So, work done by pulling force = Wp = 500*5*cos(30 deg)

Wp = 2165 J

b.)

here, F = friction force = fr = $\mu$*N

where, N = normal reaction force on sled

by force balance in vertical direction,

N + F*sin$\theta$ = m*g

N = m*g - F*sin$\theta$

given, m = mass of sled = 50 kg

$\mu$ = 0.3

$\theta$ = 30 deg

So, fr = 0.3*(50*9.8 - 500*sin(30 deg)) = 72 N

$\Theta$ = angle between friction force and displacement = 180 deg

therefore work done by friction force = Wfr = fr*d*cos$\Theta$

Wfr = 72*5*cos(180 deg)

Wfr = -360 J

c.)

From Free body diagram, angle between normal force and displacement = 90 deg

So, Work done by normal force = Wn = N*d*cos(90 deg)

Wn = 0 J

d.)

also, From Free body diagram, angle between gravity force and displacement = 90 deg

So, Work done by gravity force = Wg = (m*g)*d*cos(90 deg)

Wn = 0 J

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