Calculating Work Done on an Object with a Constant Force

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The discussion focuses on calculating the work done on a 2 kg object by a constant force of 10 N acting at an angle of 150 degrees as it moves to the position vector (2m)i - (4m)j. The initial calculation attempts to use the formula w = F x d and the Pythagorean theorem to find the distance, but the user is confused about the correct application of these concepts. The correct approach involves using the dot product of the force vector and the displacement vector to determine work done. Additionally, a related question arises about calculating work done when pushing a box up a ramp, indicating a need for clarification on work calculations in different scenarios. Understanding the vector components and proper formulas is crucial for accurate work calculations.
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a constant force of magnitude 10 N makes an angle of 150. (measured counterclockwise) with the direction of increasing x as the force ascts on the 2 kg object. how much work is done on the object by the force as the object moves from the origin to the point with position vector. (2m)i - (4m)j?

what did till now is w = f x d and force was equal to 10Cos 150 x (-4.47)
i used the pathoagorious theomrem and got the value -4.47. 2^2 + 4^2 and then found the root for that answer. but stil my answer is wrong. Could someone please explain me what i am doing wrong. Thanks a lot for the help.
 
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problem said:
a constant force of magnitude 10 N makes an angle of 150. (measured counterclockwise) with the direction of increasing x as the force ascts on the 2 kg object. how much work is done on the object by the force as the object moves from the origin to the point with position vector. (2m)i - (4m)j?

what did till now is w = f x d and force was equal to 10Cos 150 x (-4.47)
i used the pathoagorious theomrem and got the value -4.47. 2^2 + 4^2 and then found the root for that answer. but stil my answer is wrong. Could someone please explain me what i am doing wrong. Thanks a lot for the help.

W = \vec{F}\cdot \vec{\Delta r} = (10 \cos(150) \vec{i}+10 \sin(150) \vec{j})\cdot(2\vec{i}-4\vec{j}).
 
another quick question. There is a ramp at an angle now a box is being pushed up with certain force up a certain distance. The mass of the block is given. How can i figure out what the total work done is?
 
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Thread 'Variable mass system : water sprayed into a moving container'

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Thread 'Minimum mass of a block'

Thread 'Minimum mass of a block'

Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
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