As a hiker descends a hill, the work done by gravity on the hiker is A. positive

Question

As a hiker descends a hill, the work done by gravity on the hiker is A. positive and depends on the part taken. B. negative and depends on the path taken. C. positive and independent of the path taken. D. negative and independent of the path taken. 26. A 75.0 kg skier slides down a 75.0 m high slope without friction. The velocity of the silver at the bottom of the slope is A. 20.6 m/s. B. 29.7 m/s. C. 38.3 m/s. D. 40.5 m/s. E. 50.0 m/s. 27. A 200 N box is pushed up an incline that is 5.00 m long and rises 1 m. If the incline is frictionless, then the work done by the pushing force is A. 336 J. B. 305 J. C. 275 J. D. 200 J. E. 157 J. 28. A spring powered dart gun is un-stretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram projectile is placed in the gun. The kinetic energy ot the projectile when it is shot from the gun is A 0.125 J. B 0.090 J. C. 0.075 J. D. 0.051 J. E. 0.030 J.

Explanation / Answer

23) in this case the horizontal component of 120 N force is only causes the motion of the block

Fcos30o =ma

here m =mass of the block =6.00 kg , a=acceleration of the block , F= force applied on the block=120 N

120cos30o N =(6.00 kg) a

a= 17.32 m/s2

Intial velocity of the block is zeo let assume final velocity v then

v2 =2as here s=distance travelled by the block =8.00 m

v2 =2(17.32 m/s2)(8.00 m)

v =16.65 m/s

change in kinetic energy of the block

KE =KEf -KEi   here KEi =0 (intial velocity of block is zero)

KE=(1/2)mv2 =(1/2)(6.00 kg)(16.65 m/s)2 =831 J (approx)

so answer is option (C)

24)The frictional force on the car (Fr) =ma

here m is the mass of the car =2000 kg and a=acceleration of the car

from kinetmatics formula for relating acceleration,velocity and time in linear motion is

v- u =at here u=0 m/s (intially car at rest)

and t= time taken to get the velocity (v=40 m/s) =5 s

40 m/s - 0 m/s =a(5 s) and we get

a= 8 m/s2

distance(S) travelled by the car while retardation is obtained from

v2 =2as and s=v2/(2a)

s =(40 m/s)2/(2x 8 m/s2) =100 m

work done by the friction is W=Fr s

W =mas =(2000 kg)( 8 m/s2) (100 m) =1.6 x106 J

so answer is option (C)

25)the work done against the gravity is always postive . in this case force due to gravity always act downward and whatever the sign given to the force is also given to the diaplacement

work is product of force and displacement. In this case force is negative (in the direction of gravity) so displacement also negative and froduct of force and displacement always gives the postive sign

the work done by the gravity on a hiker is postive . We know that work always fallows the shortest path it is independ of path

so answer is option (C)

26)from the law of energy conservation

Energy at the top of the slope =Energy at the bottom of the slope

at the top of the slope there is only potential energy exist it is completely converted into kinetic energy at the bottom of the slope

PE(at the top of the slope) =KE (at the bottom)

mgh =(1/2) mv2 and

v= sqrt(2gh)

here g=9.8 m/s2 (acceleration due to gravity) ,h=height of the slope=75.0 m

v =sqrt(2x9.8 x75) =38.34 m/s

so answer is option (C) is answer

27)here the work done by the pushing force (F) is

W =F(5 m) (5 m is the length of the incline i.e distance travelled by the block)

and the blok is in motion when the pushing force equal to the force by the block along the incline

i.e F=mg sin(theta) here sin(theta) =rise/incline =1m/5 m =1/5

so work(W) =mg (1/5) (5) =(mg) J

here weight of the box mg is given as 200 N

work done by the pushing force is 200 J

so answer is option (D)

28)The energy stored in a compressed spring is (1/2)Kx2 which is converted into the kinetic energy of the projectile

so Kinetic energy =(1/2)Kx2

here K (spring constant) =16.0 N/m and compression in the spring (x) =8.0 cm=0.08 m

Kinetic Energy =(1/2)(16.0 N/m) (0.08 m)2 =0.051 J

so answer is option (D)

  

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