Question

show all work please.

3. A 31 kg crate full of very cute kittens is placed on an incline that is 17° below the horizontal. The crate is connected to a spring that is anchored to a vertical wall, such that the spring is parallel to the surface of the incline. 

(a) (15 points) If the crate was connected to the spring at equilibrium length, and then allowed to stretch the spring until the crate comes to rest, determine the spring constant. Assume that the incline is frictionless and that the change in length of the spring is 2.13 m.

(b) (10 points) If there is friction between the incline and the crate, would the spring stretch more, or less than if the incline is frictionless? You must use concepts pertaining to work sond energy to receive full credit.

Answer 1

31kg. mg sino LV mg ngcos 12 the componene of the force due to Gravity (remgsmo, osit) causes the mass (m331 kg) to slide along the incline. When the spring force becomes equal to mgsinit, then Party will come to rest. u s 2. kas my sin 17" kas my sin it] o xs change in the length of the spring S 2.13m (Giveny 3 ks - 31x9. 8x Sin 17° 2.13* K = 41.7 N/m f = en rohelon force) has friction : ) sustice & of the ekX mann t Y ng sing 1 t we consider the fictional fracture riction bow the shobaco, han frictional force will play the sole to oppose the sliting of the body fs UN us coefficient e treiction. V mg mgresize C & N = mgesit" Bento per te balance Now to oppose the sliding force (ie mg sin 17), hex will be preesent an med forei o extra force e frictional force (fx lin) adding to the spring force Cream Esprenk x). 2. keteen s mgsin 17° - kas ng sin 17 - uingers 17/ Hence observing two exas. O $ @ casebilly, one con conclude ini less amouent of stretching in to connected spring.

Answer 2

SOLUTION:

PART (a):

Free body diagram:

so spring mgsin17 mgcos17 mY 179

The weight of the crate (mg) can be resolved into 2 perpendicular components (mgcos$\theta$ and mgsin$\theta$) as shown in the figure.

Let be the force on the spring.

Given that the crate comes to res, hence parallel forces along the incline gets balanced.

That is:

F=mg sin 8

According to Hooke's law, we have:

F=k

Where, K is spring constant and x is elongation of spring.

Therefore, Spring constant of the spring is given by:

Kr=mg sin 8

mg sino K = 31 kg x 9.8 m/s.sin 17° 2.13 m

::K =- 88.8225 N 2.13 m

::K = 41.7 N/m

PART (b):

If there is a frictional force, there must be some energy lose due to friction.

According to Law of conservation of energy (without friction) : Work done by spring = work done by weight of the crate.

That is:

K.x² = mg sin 8 x

But if there is frictional force (F_s), the energy conservation becomes:

K2² = mg sin 8 x 1 – F,...

Therefore, the net potential energy of spring (1/2kx²) decreases and therefore, the length of spring stretch also decreases.