hi, i have 5 questions i need the complete solutions with their finalresults. is

Question

hi,

i have 5 questions i need the complete solutions with their finalresults. is there anyone could help me, i would appreciatethat.



1)A worker pushes a sled with a force of 40 N over
a level distance of 6.0 m. If a frictional force of 24 N acts onthe
wheelbarrow in a direction opposite to that of the worker, what network is
done on the wheelbarrow?



2)
A horizontal force of 200 N is applied to a
55-kg cart across a 10-m level surface. If the cart accelerates at2.0 m/s2,
then what is the work done by the force of friction as it acts toretard the
motion of the cart?



3)
If both mass and velocity of a ball are tripled,
the kinetic energy is increased by a factor of:

a) 9, b) 12, c) 3, d) 6



4)
A 1 200-kg automobile moving at 25 m/s has the
brakes applied with a deceleration of 8.0 m/s2. How far does thecar travel
before it stops?



5)
Samantha pushes a 50-N crate up a ramp 25.0 m in
length and inclined at 10° with the horizontal. What potentialenergy
change does the crate experience?



thank you in advance

Explanation / Answer

W = Fx = (40-24)N*6.0m = 96 N*m = 96J N II: Fpush = mapush = 200N =(55kg)apush apush = 200N/55kg = 3.64 m/s2 anet = 2.0 m/s2 = apush +aretd aretd = anet - apush = 2.0m/s2 - 3.64 m/s2 = -1.6 m/s2 Fretd = maretd = (55kg)(-1.6m/s2) = -88 N Wretd = Fretd(x) = (-88N)(10m) = - 880N*m The negative sign denotes that the direction of the work bythe friction is in opposition to the cart's motion. Wretd = 880 J Kinetic Energy: KE = (1/2)mv2 KE33 = (1/2)(3m)(3v)2= (1/2)(3)(9)mv2 = 27[(1/2)mv2] =27KE Kinematics: Vx = axt + vxo 0 = (-8.0m/s2)t + 25m/s t = (25m/s)/(8.0m/s2) = 3.1 s X = xo + vxot +(1/2)axt2 X = 0 + (25m/s)(3.1s)+(1/2)(-8.0m/s2)(3.1s)2 X = 77.5 m - (4*9.61m) = 39m E = KE + U + ... U = mgh F = mg m = F/g = 50N/(9.8m/s2) = 5.1 kg U = mgh U = mgh = mg(hf -hi) =(5.1kg)(9.8m/s2)[(25m)sin(10o) - 0] U = + 217N*m = + 217 J X = xo + vxot +(1/2)axt2 X = 0 + (25m/s)(3.1s)+(1/2)(-8.0m/s2)(3.1s)2 X = 77.5 m - (4*9.61m) = 39m E = KE + U + ... U = mgh F = mg m = F/g = 50N/(9.8m/s2) = 5.1 kg U = mgh U = mgh = mg(hf -hi) =(5.1kg)(9.8m/s2)[(25m)sin(10o) - 0] U = + 217N*m = + 217 J X = 77.5 m - (4*9.61m) = 39m E = KE + U + ... U = mgh F = mg m = F/g = 50N/(9.8m/s2) = 5.1 kg U = mgh U = mgh = mg(hf -hi) =(5.1kg)(9.8m/s2)[(25m)sin(10o) - 0] U = + 217N*m = + 217 J

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