Respuesta :
Answer:a) P.E= mgh
where m=40kg g=9.8m/s^2 h=2m
P.E=40×10×2
=800N
b)K.E=P.E
i.e, 1/2mv^2=mgh
1/2v^2=gh
where,g=9.8 h=2 v=?
1/2v^2=9.8×2
v^2=19.6×2
v^2=39.2
v=6.26m/s
therefore, K.E=1/2mv^
1/2×40×6.26^2
1/2×1567.5
K.E=783.8N
c) K.E=1/2mv^2
where, K.E=783.8N m=40kg v=?
783.8=1/2×40×v^2
783.8=20v^2
v^2=783.8/20
v^2=39.19
v=√39.19
v=6.26m/s^2
(a) The gravitational potential energy of the box is of 784 J.
(b) The kinetic energy of box as it strikes the ground is 783.75 J.
(c) The velocity of the box immediately before it strikes the ground is of 6.26 m/s.
Given data:
The mass of box is, m = 40 kg.
The height of shelf is, h = 2 m.
(a)
The gravitational potential energy in the given problem is nothing but the value of potential energy. Whose expression is given as,
[tex]PE = mgh[/tex]
g is the gravitational acceleration.
Solving as,
[tex]PE = 40 \times 9.8 \times 2\\\\PE = 784 \;\rm J[/tex]
Thus, the gravitational potential energy of the box is of 784 J.
(b)
Let us obtain the velocity when box falls off the shelf. So, apply the third kinematic equation of motion as,
[tex]v^{2}=u^{2}+2gh\\\\v^{2}=0^{2}+2 \times 9.8 \times 2\\\\v=\sqrt{2 \times 9.8 \times 2}\\\\v = 6.26 \;\rm m/s[/tex]
Then kinetic energy is calculated as,
[tex]KE = \dfrac{1}{2}mv^{2} \\\\KE = \dfrac{1}{2} \times 40 \times 6.26^{2} \\\\KE = 783.75 \;\rm J[/tex]
Thus, the kinetic energy of box as it strikes the ground is 783.75 J.
(c)
The velocity (v') on immediate striking of box on the ground is obtain by the kinetic energy, because energy remains conserved.
[tex]KE = \dfrac{1}{2}\times m \times v'^{2}\\\\783.75 = \dfrac{1}{2}\times 40 \times v'^{2}\\\\v' = \sqrt{783.75 \times 2/40} \\\\v'= 6.26 \;\rm m/s[/tex]
Thus, the velocity of the box immediately before it strikes the ground is of 6.26 m/s.
Learn more about the energy conservation here:
https://brainly.com/question/1603406
