Find Velocity of the wooden block at t=4 sec

[Shivam]

Homework Statement

An initially stationary wooden block with a mass of 2kg is pulled horizontally along a time dependent force f(t)i. The coefficient of static and kinetic friction between the block and the table are 0.5 and 0.25 respectively. F(t) as a function of time is shown, then find

(i) if initial velocity of the block is zero, its velocity at 4 seconds will be ?

Relevant Equations

F=ma, f=(mu)N

Here's the given F-t graph

My Attempt...
Fsmax=10N & Fk= 5N
I made equatiom F=5t-5 ( by seeing Force force and Counteracting kinetic friction)
reduced the formula to a=(5t-t)/2 wrote a=dv/dt and solved it but got wrong answer.

 

[phinds]

Problem Statement: An initially stationary wooden block with a mass of 2kg is pulled horizontally along a time dependent force f(t)i. The coefficient of static and kinetic friction between the block and the table are 0.5 and 0.25 respectively. F(t) as a function of time is shown, then find

(i) if initial velocity of the block is zero, its velocity at 4 seconds will be ?
Relevant Equations: F=ma, f=(mu)N

Here's the given F-t graph
View attachment 245621

My Attempt...
Fsmax=10N & Fk= 5N
I made equatiom F=5t-5 ( by seeing Force force and Counteracting kinetic friction)
reduced the formula to a=(5t-t)/2 wrote a=dv/dt and solved it but got wrong answer.

And since you have not shown your work, we'd just be guessing where you went wrong. The statement "wrote a=dv/dt" is NOT showing your work, it's just describing your work.

 

[kuruman]

Problem Statement: An initially stationary wooden block with a mass of 2kg is pulled horizontally along a time dependent force f(t)i. The coefficient of static and kinetic friction between the block and the table are 0.5 and 0.25 respectively. F(t) as a function of time is shown, then find

(i) if initial velocity of the block is zero, its velocity at 4 seconds will be ?
Relevant Equations: F=ma, f=(mu)N

Here's the given F-t graph
View attachment 245621

My Attempt...
Fsmax=10N & Fk= 5N
I made equatiom F=5t-5 ( by seeing Force force and Counteracting kinetic friction)
reduced the formula to a=(5t-t)/2 wrote a=dv/dt and solved it but got wrong answer.

The graph shows the applied force. Is the acceleration actually the graph shown divided by the mass? Does the block start to slide immediately after t = 0?

 

[Shivam]

The graph shows the applied force. Is the acceleration actually the graph shown divided by the mass? Does the block start to slide immediately after t = 0?

I did everything... I thought about this question for 2 hours..
I know block starts to move at t=2 sec
Also at 2 sec kinetic friction(5N) starts to act and that's why I wrote -ve5 in force equation which is reduced by kinetic friction

 

[phinds]

So I take it you are not going to show your work.

 

[Shivam]

So I take it you are not going to show your work.

Ok..sorry I'll show...just wait a minute.

 

[Shivam]

 

[Shivam]

So I take it you are not going to show your work.

I got the answer right this time ... It was calculation mistake, for all this time I thought my method was wrong... Sorry for causing trouble.
Is their is any other method can you tell me, The one you solved with

Answer is 10m/s.

 

[phinds]

Shivam, you really need to learn to use the edit button. It's a bit ridiculous to make 3 or 4 posts in a row instead of just adding to the first one.

 

[Shivam]

Shivam, you really need to learn to use the edit button. It's a bit ridiculous to make 3 or 4 posts in a row instead of just adding to the first one.

Ok..just tried the edit method... Do u have any other method to solve the problem with ?

 

[kuruman]

Ok..just tried the edit method... Do u have any other method to solve the problem with ?

This method is the one to use because you are given the variable force as a function of time. If you were given the variable force as a function of position, then using the work-energy theorem would be more direct.