Suppose you throw a small rock over a lake so that it sails over the surface of the water. The rock has a mass of 0.090 kg and is initially moving at 8.0 m/s. How much net force does it take to bring the rock to a stop in 1.00 s? A. 89 N B. 0.120 N C. 0.72 N D. 0.011 N **What do y'all think? :) thanks!!!
dan?? help, please.
ok, you need to calculate the deacceleration first using the equaions of motion.
okie! which ones are those again?
use, v=u+at plug in u = 8 v=0 t=1 and calculate a. Then multiply it with the weight of the stone to get force.
okay:) so 0=8+a(1) 0=8+a a=-8 ?
yep, now multiply it with weight.
so -8 * 0.090 kg? = -0.72 ?
would that make the answer C. 0.72 N ?
yes, the negative sign shows that you need to apply the force in opposite direction of the motion of stone.
ahh okay awesome!! thank you!!! :D do you get this one possibly as well? :/ The same net force that acts on Rock A produces three times the acceleration when it acts on Rock B. What can you say about the mass of Rock B? A. Rock B has three times the mass of Rock A. B. Rock B has one-third the mass of Rock A. C. Rock B has the same mass as Rock A. D. Rock B has a mass of zero.
not quite sure what that means :/ I think it would be either A or B? :/ since rock a is three times the acceleration? but not very sure!! :(
F=mass X acceleration, So for same force, mass is inversly proportional to acceleration. Greater the acceleration, smaller the mass.
Got it ?
okay, so would this be the correct solution? B. Rock B has one-third the mass of Rock A.
not sure if i understood correctly? haha:P
oh wait oops it should be A right?
B
Rock B has one third the mass of Rock A
ohh haha yeah makes more sense because of the greeter acceleration smaller mass right?
ahh okay thank you @Abhisar !!! :D
Yep, :)
good job everyone, i just noticed this question
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