Work done by a conservative force using change in potential energy

[Hamiltonian]

Homework Statement

Here a block of mass 1kg is placed at a point A on a rough track. If slightly pushed towards the right, it stops at point B.
we need to calculate the work done by gravity between points A and B

Relevant Equations

$W_g = -\Delta U$

we know $W_g = -\Delta U$
but here to find $\Delta U$ we will need another equation
won't it be wrong to write $$-\Delta U = -\int_1^{0.8}mgdy$$
as this equation is derived from $W_g = -\Delta U$ and as we have 2 unknowns we will need two equations.

this is a rather easy problem but I am not able to understand why we can use $\Delta U = mg\Delta h$ here as we do not know the work done by gravity.
Also since we are not given the value of $\mu$ I am not using the work energy theorem

 

[TSny]

won't it be wrong to write $$-\Delta U = -\int_1^{0.8}mgdy$$
as this equation is derived from $W_g = -\Delta U$ and as we have 2 unknowns we will need two equations.

What are the two unknowns that you are referring to here?

Also, is point B at a height of 0.8 m? In the figure, B looks to be higher than 0.8 m.

 

[Hamiltonian]

What are the two unknowns that you are referring to here?

Also, is point B at a height of 0.8 m? In the figure, B looks to be higher than 0.8 m.

the two unknowns are $\Delta U$ and $W_g$
point B is at a height 0.8m

 

[TSny]

$W_g$ is always equal to $- \Delta U$. So, if you calculate $-\Delta U$, you have $W_g$.

 

[archaic]

the two unknowns are $\Delta U$ and $W_g$
point B is at a height 0.8m

The only potential energy here is that of gravity, so $W_g=-\Delta U$.

 

[Hamiltonian]

ok think I understood my mistake