Max Accel & Force for Mass Block & Cart

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SUMMARY

The maximum acceleration of a block resting on a cart without sliding is determined by the coefficient of static friction. For a block with a mass of 1.5 kg and a cart with a mass of 3.5 kg, the maximum acceleration is calculated to be 6.86 m/s². This value is derived from the static friction force of 10.29 N, which is the product of the block's weight and the coefficient of static friction (µ = 0.7). The force required on the cart to achieve this acceleration for the entire system is 10.29 N.

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  • Understanding of Newton's Second Law (F=ma)
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  • Basic concepts of mass and acceleration
  • Ability to perform calculations involving forces and motion
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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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