Physics, 07.11.2019 05:31, ayleenmorar

Amassless spring with force constant 602 n/m is fastened at its left end to a vertical wall, as shown below. initially, the 6 kg block and 3 kg block rest on a horizontal surface with the 6 kg block in contact with the spring (but not compressing it) and with the 3 kg block in contact with the 6 kg block. the 6 kg block is then moved to the left, compressing the spring a distance of 0.6 m, and held in place while the 3 kg block remains at rest as shown below. the acceleration of gravity is 9.8 m/s2

a) determine the elastic energy u stored in the compressed spring. answer in units of j ?

b) the 6 kg block is then released and accelerates to the right, toward the 3 kg block. the surface is rough and the coefficient of friction between each block and the surface is 0.3 . the two blocks collide, stick together, and move to the right. remember that the spring is not attached to the 6 kg block. find the speed of the 6 kg block just before it collides with the 3 kg block. ? answer in units of m/s.

c) find the final speed of both blocks (stuck together) just after they collide. answer in units of m/s?

d) find the horizontal distance the blocks move before coming to rest. answer in units of m ?

Answers: 3

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Answers

answered: iddujdjddjsj8537

(A) The elastic energy stored in the compressed spring is 108.36 J.

(B) The speed of the 6 kg block just before it collides with the 3 kg block is 6.29 m/s.

(D) The distance covered by each block before coming to rest is 2.98 m

Given data:

Value of force constant is, $k=602 \;\rm N/m$ .

Mass of block 1 is, $m_{1}=6 \;\rm kg$ .

Mass of block 2 is, $m_{2}=3 \;\rm kg$ .

Spring compression distance is, $x=0.6 \;\rm m$ .

Gravitational acceleration is, $g = 9.8 \;\rm m/s^{2}$ .

Coefficient of friction between each block and the surface is, $\mu=0.3$ .

(A)

The elastic energy (U) stored in the compressed spring is,

$U=\dfrac{1}{2}kx^{2}\\U=\dfrac{1}{2}\times 602\times 0.6^{2}\\U=108.36 \;\rm J$

Thus, the elastic energy stored in the compressed spring is 108.36 J.

(B)

Apply the Work-energy concept as,

Work done by frictional force = Kinetic energy + Elastic energy

$W =KE -U\\f \times x= \dfrac{1}{2}m_{1}v^{2}-U\\(\mu \times m_{1} \times g) \times x= \dfrac{1}{2}m_{1}v^{2}-U \\(0.3 \times 6 \times 9.8) \times 0.6= \dfrac{1}{2} \times 6 \times v^{2}-108.36\\\\v^{2}=\dfrac{(0.3 \times 6 \times 9.8 \times 0.6)+108.36}{3}\\v=6.29 \;\rm m/s$

Thus, the speed of the 6 kg block just before it collides with the 3 kg block is 6.29 m/s.

(C)

Apply the conservation of linear momentum as,

Initial momentum = Final momentum

$m_{1}v+m_{2}v'=(m_{1}+m_{2})V\\$

Here, v' is the initial speed of 3 kg block, which was at rest. So, v'=0.

And V is the final speed of both blocks after the collision.

Solving as,

$6 \times 6.29 +3 \times 0=(6+3)V\\V = \dfrac{ 6 \times 6.29}{9} \\V=4.19 \;\rm m/s$

Thus, the speed of each block after the collision is 4.19 m/s.

(D)

Again use the Work energy theorem to obtain the distance (s) covered by each block before coming to rest as,

Work done by frictional force = Kinetic energy of both blocks

$W = KE_{net}\\f \times s = \dfrac{1}{2}(m_{1}+m_{2})V^{2}\\\mu \times (m_{1}+m_{2}) \times g \times s = \dfrac{1}{2}(m_{1}+m_{2})V^{2} \\\mu \times g \times s = \dfrac{1}{2} \times V^{2}\\0.3 \times 9.8 \times s = \dfrac{1}{2} \times 4,19^{2}\\s = 2.98 \;\rm m$

Thus, the distance covered by each block before coming to rest is 2.98 m.

Learn more about kinematics of motion here:

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answered: chicapunk1736

A) Ke = 108 J , B) v = 10.24 m/s , C) v₂ = 6.83 m/s and D) $x_{f}$ = 7.93 m

Explanation:

rA) let's use the elastic energy equation

Ke = ½ k x²

Let's calculate

Ke = ½ 602 0.6²

Ke = 108 J

r.b) in this case the surface has friction, so it performs a work that opposes the movement and must be equal to the variation of the mechanical energy. Let's write the mechanical energy in two points

initial maximum compression

Em₀ = Ke = ½ k x²

final. just when you touch the small block; at this point the elongation of the zero spring (x = 0)

$Em_{f}$ = K = ½ m v²

How is friction force

$W_{fr}$ = ΔDEm = $Em_{f}$ - Em₀

-fr x = ½ m v² - ½ k x²

Let's use Newton's second law to find the force of friction

fr = μ N

Equation on the y axis

N- W = 0

N = mg

fr = μ m g

Let's replace in the equation of work and energy

- (μ mg) x = ½ m v² - ½ k x²

Calculation of the speed of the large block

v² = (½ k x² - μ mg x) 2/m

v = √ (k / m x² - 2 μ g x)

v = √ (602/6 0.6² - 2 0.3 9.8 0.6) = √ (108.36 -3.53)

v = 10.24 m/s

rC) For this part we use the conservation of the moment, here we must define a system formed by the two bodies, for this system the forces during the internal crash, so that the moment is conserved

m₁ = 6 kg

m₂ = 3 kg

before the crash

p₀ = m₁ v + 0

After the crash, before starting to move

$p_{f}$ = (m₁ + m₂) v²