A person applies a force of 40 N to move a 40-kg lawnmower horizontally to cut g

Question

A person applies a force of 40 N to move a 40-kg lawnmower horizontally to cut grass over a distance of 2 m track of the lawn. The handles of the mower are inclined at an angle of 45 degree with the horizontal direction. Find the work done by the applied force. (a) 28.28 N: (b) 56.57 N: (c) 56.57 J: (d) 0: (e) none An object of 10 kg is dropped from a height of 20 m. From the conservation of the KE and PE at a height of 10 m, find the velocity at 10 m. (a) 1960 m/s, (b) 980 m/s: (c) 14 N: (d) 14 m/s: (e) none A steel wire of length 2.0 m has a diameter of 0.20 cm is used in a musical instrument. A force of 2000 N is applied to stretch the wire by 0.20 cm. Find the Young's modulus for the wire. (a) 1E11N/m^2: (b) 1 E-11 N/m^2: (c) 2.0 11 N/m^2: (d) 4.08E11 N/m^2: (e) none In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radius of r = 10.00 cm. The pressure is transmitted by an incompressible liquid to a second piston of radius 20.0 cm. What force must the compressed air exert on the small piston in order to lift a car weighing 15,000 N? (a) 60,000 N, (b) 30,000 N, (c) 7, 500 N, (d) 3, 750 N, (e) none

Explanation / Answer

1) Work done by force = 40xcos45ox2 = 56.57 J

Correct option is (c)

2)initial potential energy = mgh = 10x9.8x20 = 1960 J

Final potential energy = 10x9.8x10 = 980 J

decrease in potential energy = 1960-980 = 980 J

increase in kinetic energy = 980 J

(1/2)mv2=980

v = 14 m/s

correct option is (d)

3) Elongation of wire = 0.2cm = 0.002m

Length of wire = 2m

Strain in wire = 0.002/2 = 0.001

Area of cross section of wire =A=pi*(0.002)2/4

= 3.141x10-6 m2

tensile stress = 2000/A = 636619772 N/m2

Young's Modulus = Stress/strain = 6.366x1011 N/m2

Correct option is (e) None

4) pressure on second piston = Weight of car/Area of second piston

Equal pressure would be applied to small piston by compressed air

Load on small piston = (Weight of car/area of second piston)xarea of small piston

= [15000/pi(202/4)]x(pi(102/4))

= 3750 N

Correct option is (d)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.