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mariarad [96]
3 years ago
8

A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95

.0kg) stands on a platform 45.0m above the ground , and one endof the cord is tied securley to his ankle and the other end to theplatform. You have promised him that when he steps off he will falla maximum of 41m before the cord stops him. You had several bungeecords to select from and tested them by stretching them out, tyingone end to a tree and pulling the other end with a force of 380.0N.When you did this, what distance will the bungee cord you shouldselect have stretched?
Physics
1 answer:
madam [21]3 years ago
6 0

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as  y=l'-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m

Now the force is

F=kx\\ or

x=\dfrac{F}{k}\\

So here F=380 N, k=630.92 N/m

x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m

So the distance is 0.602 m

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The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

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