Ap Physics C 1989 M2?

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Ap Physics C 1989 M2?

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Answer 1 · 2018-06-11T21:11:23

a) This is a basic application of newton's second law.

$2 M g - T_{v} = 2 M a$ $2Mg - T_v = 2Ma$

$2 M g - 2 M a = T_{v}$ $2Mg - 2Ma = T_v$

$2 M (g - a) = T_{v}$ $2M(g - a) = T_v$

b) First of all, we know that what will make the pulley have a non zero angular acceleration is the tension.

We know torque to be $I α$ $I alpha$ and $α = \frac{a}{r}$ $alpha = a/r$ , therefore, $τ = I (\frac{a}{r})$ $tau = I(a/r)$ . Furthermore, torque is defined as $lever arm ∙ force$ $"lever arm" • "force"$ , so we can say

$T_{v} R - T_{h} R = I (\frac{a}{r})$ $T_vR - T_hR = I(a/r)$
$T_{v} R - T_{h} R = 3 M R^{2} (\frac{a}{R})$ $T_vR - T_hR = 3MR^2(a/R)$
$T_{v} - T_{h} = 3 M a$ $T_v - T_h = 3Ma$
$2 g M - 2 M a - T_{h} = 3 M a$ $2gM - 2Ma - T_h = 3Ma$
$2 g M - 5 M a = T_{h}$ $2gM - 5Ma = T_h$
$2 g M - 5 M a = T_{h}$ $2gM-5Ma = T_h$

c) Now for some linear dynamics! We know that the force of friction is given by $F_{f} = μ F_{n} = μ m_{total} g$ $F_f = mu F_n = mu m_"total" g$

Our expression for net force will therefore be

$T_{h} - F_{f} = 3 M a$ $T_h - F_f = 3Ma$
$2 M g - 5 M a - μ m_{total} g = 3 M a$ $2Mg - 5Ma - mu m_"total" g = 3Ma$

Here we will be taking the mass of the top block for the normal force , because we are considering the friction between the two blocks.

$2 M g - 5 M a - μ 4 M g = 3 M a$ $2Mg -5Ma - mu 4Mg = 3Ma$
$2 M g - 4 M g μ = 8 M a$ $2Mg - 4Mgmu = 8Ma$
$2 M g - 4 M g μ = 8 M a$ $2Mg - 4Mgmu = 8Ma$
$g - 2 g μ = 4 a$ $g - 2gmu = 4a$

We know the value of $a$ $a$ so

$g - 2 g μ = 4 a$ $g- 2gmu = 4a$
$g (1 - 2 μ) = 8$ $g(1 - 2mu) = 8$
$1 - 2 μ = \frac{8}{g}$ $1 - 2mu = 8/g$
$μ \approx 0.1$ $mu ~~ 0.1$

d) The only force acting on the top box is friction, therefore

$4 M g μ = M_{c} a_{c}$ $4Mg mu = M_c a_c$
$4 M g μ = 4 M a_{c}$ $4Mg mu = 4Ma_c$
$g (0.1) = a_{c}$ $g(0.1) = a_c$
$a_{c} \approx 1$ $a_c ~~ 1$ m/ $s^{2}$ $s^2$

This concludes this problem. Hopefully this helps!

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Ap Physics C 1989 M2?

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