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daser333 [38]
3 years ago
11

There are (one can say) three coequal theories of motion for a single particle: Newton's second law, stating that the total forc

e on an object causes its acceleration; the work–kinetic energy theorem, stating that the total work on an object causes its change in kinetic energy; and the impulse–momentum theorem, stating that the total impulse on an object causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 4.00-kg object has velocity 7.00ĵ m/s. Then, a constant net force 11.0î N acts on the object for 4.50 s.a) Calculate the object's final velocity, using the impulse–momentum theorem.vf= m/s
Physics
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

vf = 14.2176 m/s

Explanation:

Given

m = 4 Kg

viy = 7.00 ĵ m/s

Fx = 11.0 î N

t = 4.5 s

vf = ?

Using the Impulse - Momentum Theorem, we have

F*Δt = m*Δv    ⇒  F*Δt = m*(vf - vi)

⇒    vf = (F*Δt + m*vi) / m

⇒    vf = (F*Δt + m*vi) / m

For <em>x-component</em>

⇒    vfx = (Fx*Δt + m*vix) / m = (11 N*4.5 s + 4 Kg*0 m/s) / (4 Kg)

⇒    vfx = 12.375 î m/s

For <em>y-component</em>

⇒    vfy = (Fy*Δt + m*viy) / m = (0 N*4.5 s + 4 Kg*7 m/s) / (4 Kg)

⇒    vfy = 7 ĵ m/s

Finally:

vf = √(vfx² + vfy²)

⇒   vf = √((12.375 m/s)² + (7 m/s)²)

⇒   vf = 14.2176 m/s

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Explanation:

Given that,

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(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

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(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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