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Vikentia [17]
3 years ago
9

A 0.02kg dart is loaded into a toy spring gun by compressing the spring 6cm. If the spring has a constant of 20 N/m, find… a. Th

e work needed to compress the spring. (0.036 J) b. The total energy stored in the compressed spring. (0.036 J) c. The kinetic energy of the dart as it leaves the gun. (0.036 J) d. The speed of the dart as it leaves the gun. (1.90 m/s) e. The height the dart would go if shot straight up. (0.18 m)
Physics
1 answer:
katrin2010 [14]3 years ago
6 0

Answer:

a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m

Explanation:

Mass of the dart = 0.02kg, the spring was compressed to 6cm

Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m

Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m

Work needed to compress the spring = 0.036J

b) the total energy stored in the spring = the work done to compress the spring = 0.036J

c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.

d) 1/2mv^2 = 0.036

mv^2 = 0.036*2

v^2 = 0.036*2 / 0.02 = 3.6

v = √3.6 = 1.897 approx 1.9m/s

e) kinetic energy of the dart = work done against gravity to get the body to height h

Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward

0.036 = 0.02 * 9.81 * h

0.036 / ( 0.02*9.81) = h

h = 0.18 m

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P= 454.11 N

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The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

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In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

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8 0
3 years ago
A student releases a marble from the top of a ramp. The marble increases
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Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

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Therefore,

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a = (320 cm)/(4.6 s)²

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Now, we use first equation of motion:

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<u>Vf = 69.56 cm/s</u>

7 0
3 years ago
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