Question

IP A lead weight with a volume of 6.2×10⁻⁶ m3 is lowered on a fishing line into a lake to a depth of 1.0 m.

A. What tension is required in the fishing line to give the weight an upward acceleration of 2.4 m/s2 ? Express your answer using two significant figures.

B. What acceleration will the weight have if the tension in the fishing line is 1.5 N ? Give both direction and magnitude. Express your answer using two significant figures.

Answer 1

Part A.

Using Force balance on lead weight:

F_net = T + Fb - W

From Newton's 2nd law: F_net = m*a

Fb = Buoyancy Force = V*rho_w*g

T = Tension force in fishing line = ?

W = Weight of lead = m*g = rho_lead*V*g

Given that:

V = Volume of lead weight = 6.20*10^-6 m^3

rho_w = density of water = 1000 kg/m^3

g = 9.8 m/sec^2

rho_lead = 11340 kg/m^3

a = acceleration of weight = 2.4 m/sec^2

So, Now

T + Fb - W = m*a

T = m*a + W - Fb

T = rho_lead*V*a + rho_lead*V*g - rho_w*V*g

Using above values:

T = 11340*6.2*10^-6*2.4 + 11340*6.2*10^-6*9.81 - 1000*6.2*10^-6*9.81

T = 0.80 N

Part B.

Now when T = 1.5 N, then

T + Fb - W = m*a

a = (T + Fb - W)/m

a = (T + rho_w*V*g - rho_lead*V*g )/(rho_lead*V)

a = (1.5 + 1000*6.2*10^-6*9.81 - 11340*6.2*10^-6*9.81)/(11340*6.2*10^-6)

a = 12.39 m/sec^2

In two significant figures

a = 12 m/sec^2

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