Question

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is μk = 0.400. You apply a constant force F⃗ to the block. F⃗ has magnitude 88.0 N and is directed 3 toward the wall. The spring is compressed 80.0 cm. (a) What is the speed of the block? (b) What is the magnitude of the block’s acceleration? (c) What is the direction of the block’s acceleration?

Answer 1

Answer:

a) v = 0

b) The aceleration is 1.41 $m/s^{2}$

c) The block is accelerating away from the wall.

Explanation:

First, you need to think about the effect this constant force is causing in the spring: it causes a displacement in the equilibrium point of the system, therefore we need to know where it sits now:

At equilibrium no movement is present reducing friction to 0:

$\sum{F} = 0 = F_{spring} - F_{external}$

$F_{spring} = F_{external}$

$Kx = F_{external}$

$x = \frac{F_{external}}{K}=\frac{88}{130}=0.68m=68cm$

This means that the spring can be compressed with the single force up to 68 cm, Any further compression will cause an unbalanced system and the occilation of the mass.

The spring can't be compressed by the given force to 80 cm, therefore it must have been compressed by another force and then released.

In this case, the instantanous speed is 0, since the block has just been released.

In the same instant we can stimate the free body diagram of forces by the next two equations:

$\sum_y{F}={F_N-W}=0\\\sum_x{F}={F_{spring}-F_{external}-F_{friction}}=ma$

For the y axis:

$F_N = W = mg = 3*9.8 = 29.4N$

To calculate the force of friction:

$F_{friction} = \mu_k F_N=0.4*29.4 = 11.76N$

Therefore for x axis:

${Kx-F_{external}-F_{friction}}=ma$

$a = \frac{130*0.8-88-11.76}{3} = \frac{104-88-11.76}{3}=\frac{4.24}{3}=1.41\frac{m}{s^2}$