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I am going to submit one of those MasteringPhysics online assignments soon and I was wondering if anyone would be so kind as to check my answers.

http://session.masteringphysics.com/problemAsset/1010969/21/MLD_2l_1.jpg

The question is to find the tension in the upper and lower ropes when the system is at rest, and when it has acceleration [itex]a[/itex].

Case 1: The blocks are at rest. For [itex]M_1[/itex]

[itex]T_1 - T_2 - M_1g = 0[/itex]

For [itex]M_2[/itex]

[itex]T_2 - M_2g = 0[/itex]

so

[itex]T_2 = M_2g[/itex]

and

[itex]T_1 = g(M_1 + M_2)[/itex].

Case 2: The blocks are now accelerating upward. For [itex]M_1[/itex]

[itex]T_1 - T_2 - M_1g = M_1a[/itex]

For [itex]M_2[/itex]

[itex]T_2 - M_2g = M_2a[/itex]

so

[itex]T_2 = M_2(a+ g)[/itex]

and

[itex]T_1 = (a + g)(M_1 + M_2)[/itex].

http://session.masteringphysics.com/problemAsset/1010804/15/9374.jpg

For this question you need to find [itex]F[/itex] in terms of [itex]T[/itex], given that each block has mass [itex]m[/itex]. I used Newton's second law for each block to give

[itex]T_{AB} = ma[/itex] (1)

[itex]T-T_{AB} = ma[/itex] (2)

[itex]F-T = ma[/itex]

combining (1) and (2) gives

[itex]ma = \frac{1}{2}T[/itex]

so

[itex]F = \frac{3}{2}T[/itex].

http://session.masteringphysics.com/problemAsset/1010982/19/MLD_cm_7_a.jpg

A car rounding a frictionless banked curve with uniform speed [itex]v[/itex]. Find the radius [itex]r[/itex] of the turn.

I'm fairly sure the solution is [itex]r = \frac{v^2\cot\theta}{g}[/itex]

Now [itex]\theta = 0[/itex] and there is a coefficient of static friction [itex]\mu[/itex] between the road and the car's tyres. Find the mimimum value of the coefficient of static friction required to prevent the car from slipping. Assume that the car's speed is still [itex]v[/itex] and that the radius of the curve is given by [itex]r[/itex]. For this I reasoned that [itex]\frac{Mv^2}{r}[/itex] is some multiple of the normal force [itex]Mg[/itex], so [itex]\mu_\mathrm{min} = \frac{v^2}{rg}[/itex].

http://session.masteringphysics.com/problemAsset/1010934/31/MFS_1l_3_v1_a.jpg

Find [itex]T_1[/itex].

Force balance in the horizontal and vertical directions gives

[itex]T_1\cos\theta_1 = T_2\cos\theta_2[/itex] (3)

and

[itex]T_1\sin\theta_1 + T_2\sin\theta_2 = mg[/itex]. (4)

Eliminating [itex]T_2[/itex] from (4) gives

[itex]T_1\sin\theta_1 + T_1\cos\theta_1\tan\theta_2 = mg[/itex]

so

[itex]T_1 = \frac{mg}{(\sin\theta_1 + \cos\theta_1\tan\theta_2)}[/itex].

http://session.masteringphysics.com/problemAsset/1010976/22/MLD_2l_2_v2_2_a.jpg

Find the ratio of the masses [itex]m_1/m_2[/itex].

For [itex]m_1[/itex]

[itex]-m_1a = T - m_1g[/itex]

[itex]T= m_1(g - a)[/itex]

For [itex]m_2[/itex]

[itex]m_2a = T - m_2g\sin\theta - \mu m_2g\cos\theta[/itex]

[itex]m_2a = m_1(g - a) - m_2g\sin\theta - \mu m_2g\cos\theta[/itex]

[itex]m_2(a + g\sin\theta + \mu g\cos\theta) = m_1(g - a)[/itex]

[itex]m_1/m_2=\frac{a + g\sin\theta + \mu g\cos\theta}{g-a}[/itex]

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.3×10^6 N, one at an angle 17.0 deg west of north, and the other at an angle 17.0 deg east of north, as they pull the tanker a distance 0.900 toward the north. What is the total work done by the two tugboats on the supertanker?

[itex]

\begin{align*}

W & = \Sigma F_\mathrm{N}s \\

& = (2\times 1.3\times 10^6\thinspace\mathrm{N}\cos 17°)(900\thinspace

\mathrm{m}) \\

& = 2.24\times 10^9\thinspace\mathrm{J}

\end{align*}[/itex].

You are a member of an alpine rescue team and must project a box of supplies, with mass [itex]m[/itex], up an incline of constant slope angle [itex]\alpha[/itex] so that it reaches a stranded skier who is a vertical distance [itex]h[/itex] above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient [itex]\mu_\mathrm{k}[/itex]. Use the work-energy theorem to calculate the minimum speed [itex]v[/itex] that you must give the box at the bottom of the incline so that it will reach the skier.

[itex]\frac{1}{2}mv^2 - \mu_\mathrm{k}mg\cos\alpha = mgh[/itex]

[itex]v= \sqrt{2g(\mu_\mathrm{k}\cos\alpha + h)}[/itex].