Question

The spring has a constant of 29 N/m and the frictional surface is 0.4 m long with a coefficient of friction µ = 1.65. The 7 kg block depresses the spring by 19 cm, then is released. The first drop is 2.8 m and the second is 1 m. The acceleration of gravity is 9.8 m/s2 . How far from the bottom of the cliff does it land?

Answer 1

Answer:

The block lands 3 m from the bottom of the cliff.

Explanation:

Hi there!

(atteched find a figure representing the situation of the problem).

To solve this problem let´s use the theorem of conservation of energy.

Initially, the object has elastic (EPE) and gravitational potential energy (PE):

PE = m · g · h

EPE = 1/2 · k · x²

Where:

m = mass of the block.

g = acceleration due to gravity.

h = height.

k = spring constant.

x = compression of the spring.

At the bottom of the cliff, this total energy, minus some energy that will be dissipated by friction during the 0.4 m displacement over the frictional surface, will be converted into kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the block

v = velocity of the block.

The work done by friction (Wf) is equal to the dissipated energy:

Wf = Fr · d

Where:

Fr = friction force.

d = distance.

The friction force is calculated as follows:

Fr = μ · N = μ · m · g

Where:

N = normal force.

g = acceleration due to gravity.

Then, the final kinetic energy can be calculated as follows:

EPE + PE - Wf = KE

EPE = 1/2 · k · x²

EPE = 1/2 · 29 N/m · (0.19 m)²

EPE = 0.52 J

PE = m · g · h

PE = 7 kg · 9.8 m/s² · (2.8 m + 1m)

PE = 260.7 J

Wf = μ · m · g · d

Wf = 1.65 · 7 kg · 9.8 m/s² · 0.4 m

Wf = 45.3 J

Then:

KE = 0.52 J + 260.7 J - 45.3 J

KE = 215.9 J

Then, we can calculate the magnitude of the velocity when the block reaches the ground:

KE = 1/2 · m · v²

215.9 J = 1/2 · 7 kg · v²

v² = 215.9 J · 2 / 7 kg

v = 7.9 m/s

The time it takes the block to reach the ground from the second drop, can be calculated with the following equation:

h = h0 + v0y · t + 1/2 · g · t²

Where:

h = height at time t.

h0 = initial height.

v0y = initial vertical velocity.

g = acceleration due to gravity.

t = time.

When the block reaches the ground its height is zero. Initially, the block does not have vertical velocity, then, v0y = 0. The initial height is 1 m. Considering the upward direction as positive, the acceleration of gravity is negative:

h = h0 + v0y · t + 1/2 · g · t²

0 m = 1 m + 0 · t - 1/2 · 9.8 m/s² · t²

-1 m = -4.9 m/s² · t²

t² = -1 m / -4.9 m/s²

t = 0.45 s

The vertical velocity (vy), when the block reaches the ground can now be calculated:

vy = v0y + g · t

vy = -9.8 m/s² · 0.45 s

vy = -4.4 m/s

And now, we can finally find the horizontal velocity (vx) of the block. The magnitude of the velocity when the block reaches the ground is calcualted as follows:

$v = \sqrt{ vx^{2} + vy^{2} }$

v² = vx² + vy²

v² - vy² = vx²

√(v² - vy²) = vx

vx = √((7.9 m/s)² - (4.4 m/s)²)

vx = 6.6 m/s

Since there is no force accelerating the block in the horizontal direction, the horizontal velocity of the block when it lands is equal to the initial horizontal velocity. Then, we can calculate the horizontal traveled distance:

x = x0 + v · t   (x0 = 0 because we consider the edge of the cliff as the origin of the frame of reference).

x = 0 + 6.6 m/s · 0.45 s

x = 3 m

The block lands 3 m from the bottom of the cliff.