Answer:
Explanation:
given,
initial velocity of the ball = 20 m/s
angle of ramp = 22°
ball travel at a distance = 5 m
a) for friction less
![\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dmu%5E2%20-%20mgh)
![v^2 = u^2 - 2gh](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20-%202gh)
![v = \sqrt{u^2- 2 g h cos 22^0}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bu%5E2-%202%20g%20h%20cos%2022%5E0%7D)
![v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B20%5E2-%202%5Ctimes%209.8%20%5Ctimes%205%20cos%2022%5E0%20%7D)
v = 17.58 m/s
b) considering the friction
![\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dmu%5E2%20-%20mgh-%5Cmu_kmgl)
![v^2 = u^2 - 2gh-2\mu_kmgl](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20-%202gh-2%5Cmu_kmgl)
![v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bu%5E2-%202%20g%20h%20cos%2022%5E0-2%5Cmu_kgl%7D)
![v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B20%5E2-%202%5Ctimes%209.8%20%5Ctimes%205%20cos%2022%5E0-2%5Ctimes%200.15%5Ctimes%209.8%20%5Ctimes%205%20%7D)
v = 17.16 m/s
Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s
The answer would be 3,145N. Using W=mg solve for the mass of the man on earth. Once you have the mass you can multiply it by the gravity of Jupiter giving you his weight in Newton’s on Jupiter.
The correct answer is D, Diamond
Let k = the force constant of the spring (N/m).
The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
= 0.5*(k N/m)*(0.35 m)²
= 0.06125k J
The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
= 0.5*(1.8 kg)*(1.9 m/s)²
= 3.249 J
Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m
Answer: 53 N/m (nearest integer)