A 40 kg block is lifted vertically 20 meter, at the surface of the earth To thre

Question

A 40 kg block is lifted vertically 20 meter, at the surface of the earth To three significant figure, what is the change in the potential energy of the block? + 8000J -8000 J + 800 J -800 J -25600 J A shier leaves the top of a slope with an initial speed of 5.0 m/S Her speed at the bottom of the slope is 13 m/s. What is the height of the slope? 1-1 m 4.6 m 6.4 m 10 m 7.3 m A 1000-kg elevator moves upward with constant speed a vertical distance of 30.0 m. How much work was done by the tension in the cable? 980 J 30.0001 cannot be determined 1000 J 294.000 J since the speed was act specified A 10.0 kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s The crate then slides down a frictionless incline and over a second horizontal frictionless surface as shown in the figure. 30.0 J 294 J 490 J 80 J 374 J While the crate slides along the upper surface, bow much gravitational potential energy does it have with respect to the lower surface? 30 J 291 J 490 J 80 J 374 J What is the kinetic energy of the crate as it slides on the lower surface? 394 J 324 J 374 j 490 j 570 J What is the speed of the crate as it slides on the lower surf at 7.7 m/s 8.6 m/s 58.8 m/s 74.8 m/s 98 m/s

Explanation / Answer

12) m = 40 kg , h = 20 m

U2 -U1 = mgh - 0 =40*9.8*20 = 8000 J

Correct option is (a)

13) u =5 m/s, v =13 m/s

from conseration of energy

K1+U1 =K2+u2

(1/2)mu^2 +mgh = (1/2)mv^2 +0

5^2 +(2*9.8*h) = (13^2)

h = 7.3 m

correct option is (d)

14) m =1000 kg, h =30 m

W = mgh =1000*9.8*30 = 294000 J

Correct option is (d)

15) m =10 kg , v = 4 m/s

K = (1/2)mv^2 = 0.5*10*4*4 = 80 J

correct option is (b)

16) U = mgh = 10*9.8*3 = 294 J

correct option is (c)

17) From conservation of energy

K = U = 294 J

correct option is (a)

18) K =(1/2)mv^2 = 294

0.5*10*v^2 = 294

v = 7.7 m/s

correct option is (a)

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