Max Accel & Force for Mass Block & Cart

AI Thread Summary
The maximum acceleration of the block on the cart without sliding is calculated to be 6.86 m/s², derived from the static friction force of 10.29 N acting on the block. This force is obtained by multiplying the weight of the block (14.7 N) by the coefficient of static friction (0.7). To find the force required on the cart to achieve this acceleration for the entire system, the total mass (block plus cart) must be considered. The necessary force on the cart is not simply the static friction force but must account for the combined mass of both the block and the cart. The discussion clarifies that the required force on the cart is essential for achieving the maximum acceleration of the entire system.
PhysicslyDSBL
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Homework Statement



A block of mass mblock = 1.5 kg rests on a cart of mass mcart = 3.5 kg, which moves without friction on a horizontal surface. The coefficient of static friction between the block and the cart is µ = 0.7. The cart is accelerating to the right.

a) What is the maximum acceleration such that the block does not slide on the cart?


b) What is the magnitude of the force on the cart required to provide the acceleration amax?

Homework Equations



a) and b) F=ma

The Attempt at a Solution



a) I got a by taking the mass of the block and multiplying it by gravity ---> 1.5*9.8=14.7N
Then I took 14.7N and multiplied that by mew ---> 14.7*0.7 = 10.29
I then took this number, and divided it by the mass of the block --->
10.29/1.5 = 6.86 m/s^2

So 6.86m/s^2 is the max acceleration so wouldn't this make 10.29 the magnitude of the force required to provide the acceleration for amax?
 
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Hi PhysicslyDSBL,

PhysicslyDSBL said:

Homework Statement



A block of mass mblock = 1.5 kg rests on a cart of mass mcart = 3.5 kg, which moves without friction on a horizontal surface. The coefficient of static friction between the block and the cart is µ = 0.7. The cart is accelerating to the right.

a) What is the maximum acceleration such that the block does not slide on the cart?


b) What is the magnitude of the force on the cart required to provide the acceleration amax?

Homework Equations



a) and b) F=ma

The Attempt at a Solution



a) I got a by taking the mass of the block and multiplying it by gravity ---> 1.5*9.8=14.7N
Then I took 14.7N and multiplied that by mew ---> 14.7*0.7 = 10.29
I then took this number, and divided it by the mass of the block --->
10.29/1.5 = 6.86 m/s^2

So 6.86m/s^2 is the max acceleration so wouldn't this make 10.29 the magnitude of the force required to provide the acceleration for amax?

That is the required force on the cup to give it an acceleration of amax. But they want the required force on the cart to give the entire system an acceleration of amax.
 
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