Part A What is the mass of block C if block B is moving to the right and speedin

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Part A

What is the mass of block C if block B is moving to the right and speeding up with an acceleration 1.30m/s2 ?

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Part B

What is the tension in each cord when block B has this acceleration?

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Part C

Block A in the figure (Figure 1) has a mass of 4.50kg , and block B has mass 12.0kg . The coefficient of kinetic friction between block B and the horizontal surface is 0.35.

Part A

What is the mass of block C if block B is moving to the right and speeding up with an acceleration 1.30m/s2 ?

mC =   kg  

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Part B

What is the tension in each cord when block B has this acceleration?

TAB =   N  

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Part C

TBC =   N   Block A in the figure (Figure 1) has a mass of 4.50kg , and block B has mass 12.0kg . The coefficient of kinetic friction between block B and the horizontal surface is 0.35. Part A What is the mass of block C if block B is moving to the right and speeding up with an acceleration 1.30m/s2 ? Part B What is the tension in each cord when block B has this acceleration? Part C

Explanation / Answer

Apply Newton's 2nd Law ("Fnet = ma") to each of the three blocks separately; then you'll have all the equations you need and you can just use elementary algebra to get your answer.

Let mA = mass of block A (4.5kg)
mB = mass of block B (12kg)
mC = mass of block C (= ??)
a = The acceleration of Block B (1.3 m/s2)
? = coefficient of friction (0.35)
T_ab = Tension in the cord between A and B (=??)
T_bc = Tension in the cord between B and C (=??)

Notice that, since all the blocks are tied together, they are ALL accelerating at the same rate (a = 1.3 m/s2).
Now do Fnet = ma for each block separately

Block A
Downward pull (gravity): (mA)(g)
Upward pull (tension): T_ab
Net upward force: Fnet = T_ab ? (mA)(g)
So by Fnet = mAa:
T_ab ? (mA)(g) = (mA)(a)
T_ab - 4.5 * 9.8 = 4.5 * 1.3
T_ab = 49.95 N

Block B:
Leftward force (Tension): T_ab
Leftward force (friction): (mB)(g)?
Rightward force (Tension): T_bc
Net rightward force: T_bc ? T_ab ? (mB)(g)?
So by Fnet = mBa:
T_bc ? T_ab ? (mB)(g)? = (mB)(a)
T_bc ? 49.95 ? 12 * 9.8 * 0.35 = 12 * 1.3
T_bc - 49.95 - 41.16 = 15.6
T_bc = 106.71 N

Block C:
Downward pull (gravity): (mC)(g)
Upward pull (tension): T_bc
Net downward force: (mC)(g) ? T_bc
So by Fnet = mCa:
(mC)(g) ? T_bc = (mC)(a)
(mC) * 9.8 ? 106.71 = (mC) * 1.3
8.5 mC = 106.71
mC = 12.554 kg

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