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PLEASE LABEL ANSWERS AND EXPLAIN BRIEFLY!
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A) Cathy takes the bus home from work. In her hand she holds a 2.0 kg cake box, tied together with a string. As she ascends the steps into the bus, the box accelerates upward with a rate of 2.5 metres/seconds^2. What is the force exerted on the string?
B) Cathy sets the box on the seat beside her. The bus accelerates from the rest to 60.0 km/hour in 4.0 seconds, and the box begins to slide. What is the coefficient of static friction between the box and the seat of the bus?
C) A taxi suddenly cuts in front of the bus, causing the bus driver to slam on the brakes. The bus driver reduces speed from 60.0 km/h to 20.0 km/h in 1.5 seconds. Does Cathy’s cake slide forward?
PHYSICS HELP - acceleration, static friction, etc.?
A) The acceleration of the box is due to the net force on the box. This is the tension of the string ( up ) and the weight of the box ( down ). So
T - mg = ma
T - ( 2.0 kg ) ( 9.8 m/s² ) = ( 2.0 kg ) ( 2.5 m/s²)
Solve for the tension in the string T.
B) The acceleration of the bus is found from
v = vo + at
where
v = final velocity ( 60 km / h, which needs to be expressed in m / s )
vo = initial velocity ( zero )
a = acceleration ( what you're looking for )
t = time ( 4.0 s )
Solve for a. Then F = ma = μmg where μ is the coefficient of static friction.
( 2.0 kg ) ( a ) = μ ( 2.0 kg ) ( 9.8 m/s² )
Solve for the coefficient of static friction μ.
C) Same math as B). Compute the acceleration. If it exceeds the value found in B, the box slides; if it doesn't, the box stays put.
Reply:Honey it's a friday night screw homework!!!!! I'm gettin ready to go out!!!! U should come with!
But serioulsly try using those triangles like
Force
Mass | Acceleration
answer A = i think its 5 newtons
you would do 2.0 kg multiplied by 2.5 meters/seconds^2
hope i'm right.....it's been a while since i took physics
never mind i was wrong I guess i'm pretty dumb when it comes to physics
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