A 10kg block accelerates at 2ms2 across concrete. The coefficient of friction of concrete is 0.5. How much force was applied to the block?
Force pushing on a block w friction F is the force applied to the block. W is the weight of the block, or the force due to gravity. Weight is defined as W=mg where m is the mass of the block and g is the gravitational constant. FN is the normal force acting perpendicular to the contact surface. f is the force due to kinetic friction. Friction is defined as f= FN where is the coefficient of friction. To find the force due to friction, we need to find FN by applying Newton's second in the y-direction. Newton's second law is Fnet=ma where Fnet is the net force, m is the mass of the
Asked 14 days ago|2/9/2024 7:06:28 AM
Updated 14 days ago|2/9/2024 8:38:44 AM
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Force pushing on a block w friction

F is the force applied to the block.

W is the weight of the block, or the force due to gravity. Weight is defined as W=mg where m is the mass of the block and g is the gravitational constant.

FN is the normal force acting perpendicular to the contact surface.

f is the force due to kinetic friction. Friction is defined as f= FN where is the coefficient of friction.

To find the force due to friction, we need to find FN by applying Newton's second in the y-direction.

Newton's second law is Fnet=ma where Fnet is the net force, m is the mass of the block and a is the acceleration.

There are two forces in the y-direction, FN and W. There are in opposite directions, so they are subtracted. We are given m=10kg. There is no acceleration in the y-direction, so a=0.

Substituting all this information into Newton's second law gives us

Fnet=ma FN W=(10) 0 FN mg=(10) 0

FN mg=0 FN=mg

Assuming g=10m/s^2,

FN=mg=(10)(10)=100N

Now that we have FN, we can find the force due to friction. Given that =0.5 and FN=100N

f= FN=0.5 100N=50N

We now apply Newton's second law in the direction of acceleration. In this problem, that is the x-direction.

There are two forces in the direction of acceleration,the applied force F and the force due to friction f. Assuming that F is applied in the direction of acceleration and f is in the opposite direction,

Fnet=F f=F 50. The problem tells us m=10kg and a=2m/s^2. Substituting this information into Newton's second law gives us

Fnet=ma

F 50=10(2)

F 50=20 F=70N
Added 14 days ago|2/9/2024 7:30:56 AM
Rated good by Taurus28
3
[Deleted]
Added 14 days ago|2/9/2024 7:29:30 AM
Deleted by Shaznaydee29 [2/9/2024 7:31:21 AM], Flagged by Shaznaydee29 [2/9/2024 7:31:24 AM], Deleted by Shaznaydee29 [2/9/2024 7:31:35 AM]
Hi May I know why my answer is deleted here?
Added 14 days ago|2/9/2024 7:35:19 AM
Hi good day Mr. Read, your answer is wrong I'll flag it but I accidentally click the delete sorry..
Added 14 days ago|2/9/2024 8:38:44 AM
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