Question

#37 has already seen a few attempts to answer/help/assist but I’ve already got the answer. Please view the first two questions instead if you want to help. The questions are on work, energy and power.
I apologize if I'm not able to answer right away because the site is down on my side - I can't connect to the question link in order to perform such an action.

---
48) Starting from rest, a 10.0 kg suitcase slides 3.00 m down a frictionless ramp inclined at 30.0 degrees from the floor. The suitcase then slides an additional 5.00 m along the floor before coming to a stop. Determine the following:
A, the suitcase’s speed at the bottom of the ramp (I got that answer - it was 5.42 m/s)
B, the coefficient of kinetic friction
C, the change in mechanical energy due to friction

49) A light horizontal spring has a spring constant of 105 N/m. A 2.00 kg block is pressed against one end of the spring, compressing the spring 0.100 m. After the block is released, the block moves 0.250 m to the right before coming to rest. What is the coefficient of kinetic friction between the horizontal surface and the block?

37) A 215 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 30.0 cm. Calculate the following:
a, the gravitational potential energy at A relative to B
b, the particle's kinetic energy at B
c, the particle's speed at B
d, the potential energy and kinetic energy at C

Answer 1

@AllTehMaffs

Answer 2

hallo - it's all broke again!

Answer 3

Picture?

Answer 4

actually i think i got it hold on

Answer 5

very nice ^_^

Answer 6

Maybe we do not need a picture, the potential energy differences will simply be m g times the height differences. Sliding (rather than rolling) means all the potential energy changes get transformed to kinetic energy changes, (1/2) m v^2. Total PE + KE will be constant, allowing calculation where we have both.

Answer 7

@douglaswinslowcooper I meant to type this earlier but OS wouldn't let me onto the question to do so... I've solved the equation....

@AllTehMaffs ON TO THE NEXT ONE!

Answer 8

@AllTehMaffs helpp D:

Answer 9

Not sure which unanswered. You got velocity from
loss of potential energy m g h = 0.5 m v^2 kinetic energy
v=5.4 m/s

Next note energy m g h is lost to friction work F d, where d = 5 m
m g h = 147 J = 5 F, so F = 29.4 N

mu = F/W, F is friction and W is weight

mu = 29.4 N / 98 N = 0.3 coefficient of friction

Answer 10

The change in mechanical energy was just m g h.

Answer 11

Block and spring

P.E = 0.5 k x^2 = 0.5 x 105 x .01 = 0.525 J or N-m

Work = loss of that energy = force x distance

0.525 N-m= F x 0.25 m

so F = .525/.250 = 2.1 N

mu = F/weight = 2.1 / (2)(9.8) = 0.107.

Answer 12

@douglaswinslowcooper thanks for helping... but what question are you helping me with?

Answer 13

@kittiwitti1 are these questions from MasteringPhysics?

Answer 14

@HomeschoolGrad idk?

Answer 15

48)
So yeah, you already have the velocity at the bottom of the ramp, so there are a few ways to do it. @douglaswinslowcooper 's way is super slick, and basically uses the answer to the last part of the question to find the force; since you know that the work is given by
\[W = \Delta K\]
where
\[K_i = \frac{1}{2}mv^2 \rightarrow \text{where v is the v you've already found}\]
\[K_f = 0 \rightarrow \text{since there is no velocity when the object is at rest!}\]
and you can also note from conservation of energy from the top of the ramp to the bottom,
\[K_i = U = mgh\]
thus
\[W = K_f - K_i = -mgh\]
And you also know that
\[W=Fd \cos \theta \rightarrow \text{d is distance over which the force acts}
\\ \qquad \qquad \text{ and theta is the angle between the force and the object} \ \theta = 0\]
so
\[W=Fd\]
Since you know the only force acting on the object is the force of friction, then
\[F_{friction} = \frac{W}{d}\]
\[=\frac{-mgh}{d}= -\frac{mg (3m\sin 30º)}{5m}=-\frac{3mg}{10}\]
It's negative meaning it opposes the direction of motion.
Then using that knowledge, and using the magnitude of the force, the force of friction is defined to be
\[F_f = \mu N \rightarrow \text{N is the normal force of the box; N=mg}\]
so
\[\mu = \frac{F_f}{N} = \frac{ \frac{3mg}{10}}{mg} = .3\]

You could do this similarly by using kinematics to find the acceleration of the box using the velocity from the first part.
\[v^2 = v_i^2+2ad\]
\[a=-\frac{v_i^2}{2d}\]
Then summing the forces on the box, giving
\[ \sum F = F_f=F_{net}\]
\[ F_f = ma\]
\[F_f = -\frac{mv^2}{2d}=-\frac{mv^2}{2} \frac{1}{d}\]
which is cool because you can see the kinetic energy term just falls out! And we know that mv^2\2 = mgh, proving the other method is also correct! ^_^

Answer 16

49)
\[k=105 \ N/m
\\ m=2kg
\\ x=.100m
\\ d = .25m\]
So, again, just like douglaswinslow cooper said, you know that the energy imparted to the spring is
\[U_{spring}=\frac{1}{2}kx^2\]
and then the kinetic translation (although you don't need to solve for v, since you can just plug in Uspring) is
\[E_i=E_f\]
\[U_{spring} = K\]
\[\frac{1}{2}kx^2= \frac{1}{2}mv^2\]

and you know that the work done by friction negates all of the energy given to it by the spring
\[W=\Delta K = -\frac{1}{2}mv^2=-\frac{1}{2}kx^2\]
\[W=Fd\cos\theta = F_fd\]
\[-\frac{1}{2}kx^2=F_fd\]
\[F_f = -\frac{kx^2}{2d}\]
Then to find the coefficient of friction
\[\mu N=F_f\]
\[\mu N = \frac{kx^2}{2d}\]
\[ \mu = \frac{kx^2}{2dN}\]
\[=\frac{kx^2}{2dmg}\]
\[ = \frac{(105N/m)(.1m)^2}{2(.25m)(2kg)(9.81m/s^2)}\]

Answer 17

okay, I need a few days to go over this because I still have ongoing homework (as well as a quiz in 2 days)...

Answer 18

cool cool ^^